Lazy Propagation in Segment Tree - GeeksforGeeks

29 Mar, 2022
Segment tree is introduced in previous post with an example of range sum problem. We have used the same “Sum of given Range” problem to explain Lazy propagation  
How does update work in Simple Segment Tree? In the previous post, update function was called to update only a single value in array. Please note that a single value update in array may cause multiple updates in Segment Tree as there may be many segment tree nodes that have a single array element in their ranges.Below is simple logic used in previous post. 1) Start with root of segment tree. 2) If array index to be updated is not in current node’s range, then return 3) Else update current node and recur for children.Below is code taken from previous post.  
CPP
/* A recursive function to update the nodes which have the given   index in their range. The following are parameters    tree[] --> segment tree    si     -->  index of current node in segment tree.                Initial value is passed as 0.    ss and se --> Starting and ending indexes of array elements                  covered under this node of segment tree.                  Initial values passed as 0 and n-1.    i    --> index of the element to be updated. This index            is in input array.   diff --> Value to be added to all nodes which have array            index i in range */void updateValueUtil(int tree[], int ss, int se, int i,                     int diff, int si){    // Base Case: If the input index lies outside the range    // of this segment    if (i < ss || i > se)        return;      // If the input index is in range of this node, then    // update the value of the node and its children    st[si] = st[si] + diff;    if (se != ss)    {        int mid = getMid(ss, se);        updateValueUtil(st, ss, mid, i, diff, 2*si + 1);        updateValueUtil(st, mid+1, se, i, diff, 2*si + 2);    }}
What if there are updates on a range of array indexes? For example add 10 to all values at indexes from 2 to 7 in array. The above update has to be called for every index from 2 to 7. We can avoid multiple calls by writing a function updateRange() that updates nodes accordingly. 
CPP
/* Function to update segment tree for range update in inputarray.    si -> index of current node in segment tree    ss and se -> Starting and ending indexes of elements for                which current nodes stores sum.    us and ue -> starting and ending indexes of update query    diff -> which we need to add in the range us to ue */void updateRangeUtil(int si, int ss, int se, int us,                    int ue, int diff){    // out of range    if (ss>se || ss>ue || se<us)        return ;     // Current node is a leaf node    if (ss==se)    {        // Add the difference to current node        tree[si] += diff;        return;    }     // If not a leaf node, recur for children.    int mid = (ss+se)/2;    updateRangeUtil(si*2+1, ss, mid, us, ue, diff);    updateRangeUtil(si*2+2, mid+1, se, us, ue, diff);     // Use the result of children calls to update this    // node    tree[si] = tree[si*2+1] + tree[si*2+2];}
Lazy Propagation – An optimization to make range updates fasterWhen there are many updates and updates are done on a range, we can postpone some updates (avoid recursive calls in update) and do those updates only when required.Please remember that a node in segment tree stores or represents result of a query for a range of indexes. And if this node’s range lies within the update operation range, then all descendants of the node must also be updated. For example consider the node with value 27 in above diagram, this node stores sum of values at indexes from 3 to 5. If our update query is for range 2 to 5, then we need to update this node and all descendants of this node. With Lazy propagation, we update only node with value 27 and postpone updates to its children by storing this update information in separate nodes called lazy nodes or values. We create an array lazy[] which represents lazy node. Size of lazy[] is same as array that represents segment tree, which is tree[] in below code.The idea is to initialize all elements of lazy[] as 0. A value 0 in lazy[i] indicates that there are no pending updates on node i in segment tree. A non-zero value of lazy[i] means that this amount needs to be added to node i in segment tree before making any query to the node.Below is modified update method. 
// To update segment tree for change in array
// values at array indexes from us to ue.
updateRange(us, ue)
1) If current segment tree node has any pending
   update, then first add that pending update to
   current node.
2) If current node's range lies completely in 
   update query range.
....a) Update current node
....b) Postpone updates to children by setting 
       lazy value for children nodes.
3) If current node's range overlaps with update 
   range, follow the same approach as above simple
   update.
...a) Recur for left and right children.
...b) Update current node using results of left 
      and right calls.
Is there any change in Query Function also? Since we have changed update to postpone its operations, there may be problems if a query is made to a node that is yet to be updated. So we need to update our query method also which is getSumUtil in previous post. The getSumUtil() now first checks if there is a pending update and if there is, then updates the node. Once it makes sure that pending update is done, it works same as the previous getSumUtil().Below are programs to demonstrate working of Lazy Propagation.
C++
Java
Python3
C#
// Program to show segment tree to demonstrate lazy// propagation#include <stdio.h>#include <math.h>#define MAX 1000 // Ideally, we should not use global variables and large// constant-sized arrays, we have done it here for simplicity.int tree[MAX] = {0};  // To store segment treeint lazy[MAX] = {0};  // To store pending updates /*  si -> index of current node in segment tree    ss and se -> Starting and ending indexes of elements for                 which current nodes stores sum.    us and ue -> starting and ending indexes of update query    diff -> which we need to add in the range us to ue */void updateRangeUtil(int si, int ss, int se, int us,                     int ue, int diff){    // If lazy value is non-zero for current node of segment    // tree, then there are some pending updates. So we need    // to make sure that the pending updates are done before    // making new updates. Because this value may be used by    // parent after recursive calls (See last line of this    // function)    if (lazy[si] != 0)    {        // Make pending updates using value stored in lazy        // nodes        tree[si] += (se-ss+1)*lazy[si];         // checking if it is not leaf node because if        // it is leaf node then we cannot go further        if (ss != se)        {            // We can postpone updating children we don't            // need their new values now.            // Since we are not yet updating children of si,            // we need to set lazy flags for the children            lazy[si*2 + 1]   += lazy[si];            lazy[si*2 + 2]   += lazy[si];        }         // Set the lazy value for current node as 0 as it        // has been updated        lazy[si] = 0;    }     // out of range    if (ss>se || ss>ue || se<us)        return ;     // Current segment is fully in range    if (ss>=us && se<=ue)    {        // Add the difference to current node        tree[si] += (se-ss+1)*diff;         // same logic for checking leaf node or not        if (ss != se)        {            // This is where we store values in lazy nodes,            // rather than updating the segment tree itself            // Since we don't need these updated values now            // we postpone updates by storing values in lazy[]            lazy[si*2 + 1]   += diff;            lazy[si*2 + 2]   += diff;        }        return;    }     // If not completely in rang, but overlaps, recur for    // children,    int mid = (ss+se)/2;    updateRangeUtil(si*2+1, ss, mid, us, ue, diff);    updateRangeUtil(si*2+2, mid+1, se, us, ue, diff);     // And use the result of children calls to update this    // node    tree[si] = tree[si*2+1] + tree[si*2+2];} // Function to update a range of values in segment// tree/*  us and eu -> starting and ending indexes of update query    ue  -> ending index of update query    diff -> which we need to add in the range us to ue */void updateRange(int n, int us, int ue, int diff){   updateRangeUtil(0, 0, n-1, us, ue, diff);}  /*  A recursive function to get the sum of values in given    range of the array. The following are parameters for    this function.    si --> Index of current node in the segment tree.           Initially 0 is passed as root is always at'           index 0    ss & se  --> Starting and ending indexes of the                 segment represented by current node,                 i.e., tree[si]    qs & qe  --> Starting and ending indexes of query                 range */int getSumUtil(int ss, int se, int qs, int qe, int si){    // If lazy flag is set for current node of segment tree,    // then there are some pending updates. So we need to    // make sure that the pending updates are done before    // processing the sub sum query    if (lazy[si] != 0)    {        // Make pending updates to this node. Note that this        // node represents sum of elements in arr[ss..se] and        // all these elements must be increased by lazy[si]        tree[si] += (se-ss+1)*lazy[si];         // checking if it is not leaf node because if        // it is leaf node then we cannot go further        if (ss != se)        {            // Since we are not yet updating children os si,            // we need to set lazy values for the children            lazy[si*2+1] += lazy[si];            lazy[si*2+2] += lazy[si];        }         // unset the lazy value for current node as it has        // been updated        lazy[si] = 0;    }     // Out of range    if (ss>se || ss>qe || se<qs)        return 0;     // At this point we are sure that pending lazy updates    // are done for current node. So we can return value    // (same as it was for query in our previous post)     // If this segment lies in range    if (ss>=qs && se<=qe)        return tree[si];     // If a part of this segment overlaps with the given    // range    int mid = (ss + se)/2;    return getSumUtil(ss, mid, qs, qe, 2*si+1) +           getSumUtil(mid+1, se, qs, qe, 2*si+2);} // Return sum of elements in range from index qs (query// start) to qe (query end).  It mainly uses getSumUtil()int getSum(int n, int qs, int qe){    // Check for erroneous input values    if (qs < 0 || qe > n-1 || qs > qe)    {        printf("Invalid Input");        return -1;    }     return getSumUtil(0, n-1, qs, qe, 0);} // A recursive function that constructs Segment Tree for//  array[ss..se]. si is index of current node in segment// tree st.void constructSTUtil(int arr[], int ss, int se, int si){    // out of range as ss can never be greater than se    if (ss > se)        return ;     // If there is one element in array, store it in    // current node of segment tree and return    if (ss == se)    {        tree[si] = arr[ss];        return;    }     // If there are more than one elements, then recur    // for left and right subtrees and store the sum    // of values in this node    int mid = (ss + se)/2;    constructSTUtil(arr, ss, mid, si*2+1);    constructSTUtil(arr, mid+1, se, si*2+2);     tree[si] = tree[si*2 + 1] + tree[si*2 + 2];} /* Function to construct segment tree from given array.   This function allocates memory for segment tree and   calls constructSTUtil() to fill the allocated memory */void constructST(int arr[], int n){    // Fill the allocated memory st    constructSTUtil(arr, 0, n-1, 0);}  // Driver program to test above functionsint main(){    int arr[] = {1, 3, 5, 7, 9, 11};    int n = sizeof(arr)/sizeof(arr[0]);     // Build segment tree from given array    constructST(arr, n);     // Print sum of values in array from index 1 to 3    printf("Sum of values in given range = %d\n",           getSum(n, 1, 3));     // Add 10 to all nodes at indexes from 1 to 5.    updateRange(n, 1, 5, 10);     // Find sum after the value is updated    printf("Updated sum of values in given range = %d\n",            getSum( n, 1, 3));     return 0;}
// Java program to demonstrate lazy propagation in segment treeclass LazySegmentTree{    final int MAX = 1000;     // Max tree size    int tree[] = new int[MAX]; // To store segment tree    int lazy[] = new int[MAX]; // To store pending updates     /* si -> index of current node in segment tree        ss and se -> Starting and ending indexes of elements for                    which current nodes stores sum.        us and eu -> starting and ending indexes of update query        ue -> ending index of update query        diff -> which we need to add in the range us to ue */    void updateRangeUtil(int si, int ss, int se, int us,                        int ue, int diff)    {        // If lazy value is non-zero for current node of segment        // tree, then there are some pending updates. So we need        // to make sure that the pending updates are done before        // making new updates. Because this value may be used by        // parent after recursive calls (See last line of this        // function)        if (lazy[si] != 0)        {            // Make pending updates using value stored in lazy            // nodes            tree[si] += (se - ss + 1) * lazy[si];             // checking if it is not leaf node because if            // it is leaf node then we cannot go further            if (ss != se)            {                // We can postpone updating children we don't                // need their new values now.                // Since we are not yet updating children of si,                // we need to set lazy flags for the children                lazy[si * 2 + 1] += lazy[si];                lazy[si * 2 + 2] += lazy[si];            }             // Set the lazy value for current node as 0 as it            // has been updated            lazy[si] = 0;        }         // out of range        if (ss > se || ss > ue || se < us)            return;         // Current segment is fully in range        if (ss >= us && se <= ue)        {            // Add the difference to current node            tree[si] += (se - ss + 1) * diff;             // same logic for checking leaf node or not            if (ss != se)            {                // This is where we store values in lazy nodes,                // rather than updating the segment tree itself                // Since we don't need these updated values now                // we postpone updates by storing values in lazy[]                lazy[si * 2 + 1] += diff;                lazy[si * 2 + 2] += diff;            }            return;        }         // If not completely in rang, but overlaps, recur for        // children,        int mid = (ss + se) / 2;        updateRangeUtil(si * 2 + 1, ss, mid, us, ue, diff);        updateRangeUtil(si * 2 + 2, mid + 1, se, us, ue, diff);         // And use the result of children calls to update this        // node        tree[si] = tree[si * 2 + 1] + tree[si * 2 + 2];    }     // Function to update a range of values in segment    // tree    /* us and eu -> starting and ending indexes of update query        ue -> ending index of update query        diff -> which we need to add in the range us to ue */    void updateRange(int n, int us, int ue, int diff) {        updateRangeUtil(0, 0, n - 1, us, ue, diff);    }     /* A recursive function to get the sum of values in given        range of the array. The following are parameters for        this function.        si --> Index of current node in the segment tree.            Initially 0 is passed as root is always at'            index 0        ss & se --> Starting and ending indexes of the                    segment represented by current node,                    i.e., tree[si]        qs & qe --> Starting and ending indexes of query                    range */    int getSumUtil(int ss, int se, int qs, int qe, int si)    {        // If lazy flag is set for current node of segment tree,        // then there are some pending updates. So we need to        // make sure that the pending updates are done before        // processing the sub sum query        if (lazy[si] != 0)        {            // Make pending updates to this node. Note that this            // node represents sum of elements in arr[ss..se] and            // all these elements must be increased by lazy[si]            tree[si] += (se - ss + 1) * lazy[si];             // checking if it is not leaf node because if            // it is leaf node then we cannot go further            if (ss != se)            {                // Since we are not yet updating children os si,                // we need to set lazy values for the children                lazy[si * 2 + 1] += lazy[si];                lazy[si * 2 + 2] += lazy[si];            }             // unset the lazy value for current node as it has            // been updated            lazy[si] = 0;        }         // Out of range        if (ss > se || ss > qe || se < qs)            return 0;         // At this point sure, pending lazy updates are done        // for current node. So we can return value (same as        // was for query in our previous post)         // If this segment lies in range        if (ss >= qs && se <= qe)            return tree[si];         // If a part of this segment overlaps with the given        // range        int mid = (ss + se) / 2;        return getSumUtil(ss, mid, qs, qe, 2 * si + 1) +            getSumUtil(mid + 1, se, qs, qe, 2 * si + 2);    }     // Return sum of elements in range from index qs (query    // start) to qe (query end). It mainly uses getSumUtil()    int getSum(int n, int qs, int qe)    {        // Check for erroneous input values        if (qs < 0 || qe > n - 1 || qs > qe)        {            System.out.println("Invalid Input");            return -1;        }         return getSumUtil(0, n - 1, qs, qe, 0);    }     /* A recursive function that constructs Segment Tree for    array[ss..se]. si is index of current node in segment    tree st. */    void constructSTUtil(int arr[], int ss, int se, int si)    {        // out of range as ss can never be greater than se        if (ss > se)            return;         /* If there is one element in array, store it in        current node of segment tree and return */        if (ss == se)        {            tree[si] = arr[ss];            return;        }         /* If there are more than one elements, then recur        for left and right subtrees and store the sum        of values in this node */        int mid = (ss + se) / 2;        constructSTUtil(arr, ss, mid, si * 2 + 1);        constructSTUtil(arr, mid + 1, se, si * 2 + 2);         tree[si] = tree[si * 2 + 1] + tree[si * 2 + 2];    }     /* Function to construct segment tree from given array.    This function allocates memory for segment tree and    calls constructSTUtil() to fill the allocated memory */    void constructST(int arr[], int n)    {        // Fill the allocated memory st        constructSTUtil(arr, 0, n - 1, 0);    }      // Driver program to test above functions    public static void main(String args[])    {        int arr[] = {1, 3, 5, 7, 9, 11};        int n = arr.length;        LazySegmentTree tree = new LazySegmentTree();         // Build segment tree from given array        tree.constructST(arr, n);         // Print sum of values in array from index 1 to 3        System.out.println("Sum of values in given range = " +                        tree.getSum(n, 1, 3));         // Add 10 to all nodes at indexes from 1 to 5.        tree.updateRange(n, 1, 5, 10);         // Find sum after the value is updated        System.out.println("Updated sum of values in given range = " +                        tree.getSum(n, 1, 3));    }}// This Code is contributed by Ankur Narain Verma
# Python3 implementation of the approachMAX = 1000 # Ideally, we should not use global variables# and large constant-sized arrays, we have# done it here for simplicity.tree = [0] * MAX; # To store segment treelazy = [0] * MAX; # To store pending updates """ si -> index of current node in segment tree    ss and se -> Starting and ending indexes of elements                for which current nodes stores sum.    us and ue -> starting and ending indexes of update query    diff -> which we need to add in the range us to ue """def updateRangeUtil(si, ss, se, us, ue, diff) :     # If lazy value is non-zero for current node    # of segment tree, then there are some    # pending updates. So we need to make sure    # that the pending updates are done before    # making new updates. Because this value may be    # used by parent after recursive calls    # (See last line of this function)    if (lazy[si] != 0) :                 # Make pending updates using value        # stored in lazy nodes        tree[si] += (se - ss + 1) * lazy[si];         # checking if it is not leaf node because if        # it is leaf node then we cannot go further        if (ss != se) :                     # We can postpone updating children we don't            # need their new values now.            # Since we are not yet updating children of si,            # we need to set lazy flags for the children            lazy[si * 2 + 1] += lazy[si];            lazy[si * 2 + 2] += lazy[si];                 # Set the lazy value for current node        # as 0 as it has been updated        lazy[si] = 0;         # out of range    if (ss > se or ss > ue or se < us) :        return ;     # Current segment is fully in range    if (ss >= us and se <= ue) :                 # Add the difference to current node        tree[si] += (se - ss + 1) * diff;         # same logic for checking leaf node or not        if (ss != se) :                     # This is where we store values in lazy nodes,            # rather than updating the segment tree itself            # Since we don't need these updated values now            # we postpone updates by storing values in lazy[]            lazy[si * 2 + 1] += diff;            lazy[si * 2 + 2] += diff;                 return;     # If not completely in rang, but overlaps,    # recur for children,    mid = (ss + se) // 2;    updateRangeUtil(si * 2 + 1, ss,                    mid, us, ue, diff);    updateRangeUtil(si * 2 + 2, mid + 1,                    se, us, ue, diff);     # And use the result of children calls    # to update this node    tree[si] = tree[si * 2 + 1] + \            tree[si * 2 + 2]; # Function to update a range of values# in segment tree ''' us and eu -> starting and ending indexes                of update query    ue -> ending index of update query    diff -> which we need to add in the range us to ue '''def updateRange(n, us, ue, diff) :    updateRangeUtil(0, 0, n - 1, us, ue, diff); ''' A recursive function to get the sum of values    in given range of the array. The following are    parameters for this function.    si --> Index of current node in the segment tree.        Initially 0 is passed as root is always at'        index 0    ss & se --> Starting and ending indexes of the                segment represented by current node,                i.e., tree[si]    qs & qe --> Starting and ending indexes of query                range '''def getSumUtil(ss, se, qs, qe, si) :     # If lazy flag is set for current node    # of segment tree, then there are    # some pending updates. So we need to    # make sure that the pending updates are    # done before processing the sub sum query    if (lazy[si] != 0) :             # Make pending updates to this node.        # Note that this node represents sum of        # elements in arr[ss..se] and all these        # elements must be increased by lazy[si]        tree[si] += (se - ss + 1) * lazy[si];         # checking if it is not leaf node because if        # it is leaf node then we cannot go further        if (ss != se) :                     # Since we are not yet updating children os si,            # we need to set lazy values for the children            lazy[si * 2 + 1] += lazy[si];            lazy[si * 2 + 2] += lazy[si];         # unset the lazy value for current node        # as it has been updated        lazy[si] = 0;     # Out of range    if (ss > se or ss > qe or se < qs) :        return 0;     # At this point we are sure that    # pending lazy updates are done for    # current node. So we can return value    # (same as it was for query in our previous post)     # If this segment lies in range    if (ss >= qs and se <= qe) :        return tree[si];     # If a part of this segment overlaps    # with the given range    mid = (ss + se) // 2;    return (getSumUtil(ss, mid, qs, qe, 2 * si + 1) +            getSumUtil(mid + 1, se, qs, qe, 2 * si + 2)); # Return sum of elements in range from# index qs (query start) to qe (query end).# It mainly uses getSumUtil()def getSum(n, qs, qe) :         # Check for erroneous input values    if (qs < 0 or qe > n - 1 or qs > qe) :        print("Invalid Input");        return -1;     return getSumUtil(0, n - 1, qs, qe, 0); # A recursive function that constructs# Segment Tree for array[ss..se].# si is index of current node in segment# tree st.def constructSTUtil(arr, ss, se, si) :     # out of range as ss can never be    # greater than se    if (ss > se) :        return ;     # If there is one element in array,    # store it in current node of    # segment tree and return    if (ss == se) :             tree[si] = arr[ss];        return;         # If there are more than one elements,    # then recur for left and right subtrees    # and store the sum of values in this node    mid = (ss + se) // 2;    constructSTUtil(arr, ss, mid, si * 2 + 1);    constructSTUtil(arr, mid + 1, se, si * 2 + 2);     tree[si] = tree[si * 2 + 1] + tree[si * 2 + 2]; ''' Function to construct segment treefrom given array. This function allocates memoryfor segment tree and calls constructSTUtil()to fill the allocated memory '''def constructST(arr, n) :         # Fill the allocated memory st    constructSTUtil(arr, 0, n - 1, 0);     # Driver Codeif __name__ == "__main__" :     arr = [1, 3, 5, 7, 9, 11];    n = len(arr);     # Build segment tree from given array    constructST(arr, n);     # Print sum of values in array from index 1 to 3    print("Sum of values in given range =",                        getSum(n, 1, 3));     # Add 10 to all nodes at indexes from 1 to 5.    updateRange(n, 1, 5, 10);     # Find sum after the value is updated    print("Updated sum of values in given range =",                                getSum( n, 1, 3)); # This code is contributed by AnkitRai01
// C# program to demonstrate lazy// propagation in segment treeusing System; public class LazySegmentTree{    static readonly int MAX = 1000; // Max tree size    int []tree = new int[MAX]; // To store segment tree    int []lazy = new int[MAX]; // To store pending updates     /* si -> index of current node in segment tree        ss and se -> Starting and ending indexes of elements for                    which current nodes stores sum.        us and eu -> starting and ending indexes of update query        ue -> ending index of update query        diff -> which we need to add in the range us to ue */    void updateRangeUtil(int si, int ss, int se, int us,                        int ue, int diff)    {        // If lazy value is non-zero        // for current node of segment        // tree, then there are some        // pending updates. So we need        // to make sure that the pending        // updates are done before making        // new updates. Because this        // value may be used by parent        // after recursive calls (See last        // line of this function)        if (lazy[si] != 0)        {            // Make pending updates using value            // stored in lazy nodes            tree[si] += (se - ss + 1) * lazy[si];             // checking if it is not leaf node because if            // it is leaf node then we cannot go further            if (ss != se)            {                // We can postpone updating children                // we don't need their new values now.                // Since we are not yet updating children of si,                // we need to set lazy flags for the children                lazy[si * 2 + 1] += lazy[si];                lazy[si * 2 + 2] += lazy[si];            }             // Set the lazy value for current node            // as 0 as it has been updated            lazy[si] = 0;        }         // out of range        if (ss > se || ss > ue || se < us)            return;         // Current segment is fully in range        if (ss >= us && se <= ue)        {            // Add the difference to current node            tree[si] += (se - ss + 1) * diff;             // same logic for checking leaf node or not            if (ss != se)            {                // This is where we store values in lazy nodes,                // rather than updating the segment tree itself                // Since we don't need these updated values now                // we postpone updates by storing values in lazy[]                lazy[si * 2 + 1] += diff;                lazy[si * 2 + 2] += diff;            }            return;        }         // If not completely in rang, but        // overlaps, recur for children,        int mid = (ss + se) / 2;        updateRangeUtil(si * 2 + 1, ss, mid, us, ue, diff);        updateRangeUtil(si * 2 + 2, mid + 1, se, us, ue, diff);         // And use the result of children calls to update this        // node        tree[si] = tree[si * 2 + 1] + tree[si * 2 + 2];    }     // Function to update a range of values in segment    // tree    /* us and eu -> starting and ending indexes of update query        ue -> ending index of update query        diff -> which we need to add in the range us to ue */    void updateRange(int n, int us, int ue, int diff)    {        updateRangeUtil(0, 0, n - 1, us, ue, diff);    }     /* A recursive function to get the sum of values in given        range of the array. The following are parameters for        this function.        si --> Index of current node in the segment tree.            Initially 0 is passed as root is always at'            index 0        ss & se --> Starting and ending indexes of the                    segment represented by current node,                    i.e., tree[si]        qs & qe --> Starting and ending indexes of query                    range */    int getSumUtil(int ss, int se, int qs,                            int qe, int si)    {        // If lazy flag is set for current node        // of segment tree, then there are        // some pending updates. So we need to        // make sure that the pending updates        // are done before processing        // the sub sum query        if (lazy[si] != 0)        {            // Make pending updates to this            // node. Note that this node            // represents sum of elements            // in arr[ss..se] and all these            // elements must be increased by lazy[si]            tree[si] += (se - ss + 1) * lazy[si];             // checking if it is not leaf node because if            // it is leaf node then we cannot go further            if (ss != se)            {                // Since we are not yet                // updating children os si,                // we need to set lazy values                // for the children                lazy[si * 2 + 1] += lazy[si];                lazy[si * 2 + 2] += lazy[si];            }             // unset the lazy value for current            // node as it has been updated            lazy[si] = 0;        }         // Out of range        if (ss > se || ss > qe || se < qs)            return 0;         // At this point sure, pending lazy updates are done        // for current node. So we can return value (same as        // was for query in our previous post)         // If this segment lies in range        if (ss >= qs && se <= qe)            return tree[si];         // If a part of this segment overlaps        // with the given range        int mid = (ss + se) / 2;        return getSumUtil(ss, mid, qs, qe, 2 * si + 1) +            getSumUtil(mid + 1, se, qs, qe, 2 * si + 2);    }     // Return sum of elements in range from index qs (query    // start) to qe (query end). It mainly uses getSumUtil()    int getSum(int n, int qs, int qe)    {        // Check for erroneous input values        if (qs < 0 || qe > n - 1 || qs > qe)        {            Console.WriteLine("Invalid Input");            return -1;        }         return getSumUtil(0, n - 1, qs, qe, 0);    }     /* A recursive function that constructs    Segment Tree for array[ss..se]. si is    index of current node in segment    tree st. */    void constructSTUtil(int []arr, int ss, int se, int si)    {        // out of range as ss can        // never be greater than se        if (ss > se)            return;         /* If there is one element in array, store it in        current node of segment tree and return */        if (ss == se)        {            tree[si] = arr[ss];            return;        }         /* If there are more than one elements, then recur        for left and right subtrees and store the sum        of values in this node */        int mid = (ss + se) / 2;        constructSTUtil(arr, ss, mid, si * 2 + 1);        constructSTUtil(arr, mid + 1, se, si * 2 + 2);         tree[si] = tree[si * 2 + 1] + tree[si * 2 + 2];    }     /* Function to construct segment tree from given array.    This function allocates memory for segment tree and    calls constructSTUtil() to fill the allocated memory */    void constructST(int []arr, int n)    {        // Fill the allocated memory st        constructSTUtil(arr, 0, n - 1, 0);    }      // Driver program to test above functions    public static void Main(String []args)    {        int []arr = {1, 3, 5, 7, 9, 11};        int n = arr.Length;        LazySegmentTree tree = new LazySegmentTree();         // Build segment tree from given array        tree.constructST(arr, n);         // Print sum of values in array from index 1 to 3        Console.WriteLine("Sum of values in given range = " +                        tree.getSum(n, 1, 3));         // Add 10 to all nodes at indexes from 1 to 5.        tree.updateRange(n, 1, 5, 10);         // Find sum after the value is updated        Console.WriteLine("Updated sum of values in given range = " +                        tree.getSum(n, 1, 3));    }} // This code contributed by Rajput-Ji
Output:
Sum of values in given range = 15
Updated sum of values in given range = 45 
Time Complexity: O(n)
Auxiliary Space: O(MAX) 
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This article is contributed by Ankit Mittal. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above 
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