Maximum Sum SubArray using Divide and Conquer | Set 2

10 Nov, 2021
Given an array arr[] of integers, the task is to find the maximum sum sub-array among all the possible sub-arrays.Examples:  
Input: arr[] = {-2, 1, -3, 4, -1, 2, 1, -5, 4} Output: 6 {4, -1, 2, 1} is the required sub-array.Input: arr[] = {2, 2, -2} Output: 4  
 
Approach: Till now we are only aware of Kadane’s Algorithm which solves this problem in O(n) using dynamic programming. We had also discussed a divide and conquer approach for maximum sum subarray in O(N*logN) time complexity.The following approach solves it using Divide and Conquer approach which takes the same time complexity of O(n).Divide and conquer algorithms generally involves dividing the problem into sub-problems and conquering them separately. For this problem we maintain a structure (in cpp) or class(in java or python) which stores the following values:  
Total sum for a sub-array.Maximum prefix sum for a sub-array.Maximum suffix sum for a sub-array.Overall maximum sum for a sub-array.(This contains the max sum for a sub-array).
Total sum for a sub-array.
Maximum prefix sum for a sub-array.
Maximum suffix sum for a sub-array.
Overall maximum sum for a sub-array.(This contains the max sum for a sub-array).
During the recursion(Divide part) the array is divided into 2 parts from the middle. The left node structure contains all the above values for the left part of array and the right node structure contains all the above values. Having both the nodes, now we can merge the two nodes by computing all the values for resulting node. The max prefix sum for the resulting node will be maximum value among the maximum prefix sum of left node or left node sum + max prefix sum of right node or total sum of both the nodes (which is possible for an array with all positive values). Similarly the max suffix sum for the resulting node will be maximum value among the maximum suffix sum of right node or right node sum + max suffix sum of left node or total sum of both the nodes (which is again possible for an array with all positive values). The total sum for the resulting node is the sum of both left node and right node sum. Now, the max subarray sum for the resulting node will be maximum among prefix sum of resulting node, suffix sum of resulting node, total sum of resulting node, maximum sum of left node, maximum sum of right node, sum of maximum suffix sum of left node and maximum prefix sum of right node. Here the conquer part can be done in O(1) time by combining the result from the left and right node structures.Below is the implementation of the above approach: 
C++
Java
Python3
C#
Javascript
// C++ implementation of the approach#include<bits/stdc++.h>using namespace std; struct Node {         // To store the maximum sum    // for a sub-array    long long _max;         // To store the maximum prefix    // sum for a sub-array    long long _pre;         // To store the maximum suffix    // sum for a sub-array    long long _suf;         // To store the total sum    // for a sub-array    long long _sum; };  // Function to create a nodeNode getNode(long long x){    Node a;    a._max = x;    a._pre = x;    a._suf = x;    a._sum = x;    return a;} // Function to merge the 2 nodes left and rightNode merg(const Node &l, const Node &r){         // Creating node ans    Node ans ;     // Initializing all the variables:    ans._max = ans._pre = ans._suf = ans._sum = 0;         // The max prefix sum of ans Node is maximum of    // a) max prefix sum of left Node    // b) sum of left Node + max prefix sum of right Node    // c) sum of left Node + sum of right Node    ans._pre = max({l._pre, l._sum+r._pre, l._sum+r._sum});     // The max suffix sum of ans Node is maximum of    // a) max suffix sum of right Node    // b) sum of right Node + max suffix sum of left Node    // c) sum of left Node + sum of right Node    ans._suf = max({r._suf, r._sum+l._suf, l._sum+r._sum});         // Total sum of ans Node = total sum of left Node + total sum of right Node    ans._sum = l._sum + r._sum;         // The max sum of ans Node stores the answer which is the maximum value among:    // prefix sum of ans Node    // suffix sum of ans Node    // maximum value of left Node    // maximum value of right Node    // prefix value of right Node + suffix value of left Node    ans._max = max({ans._pre, ans._suf, ans._sum,l._max, r._max, l._suf+r._pre});     // Return the ans Node    return ans;} // Function for calculating the// max_sum_subArray using divide and conquerNode getMaxSumSubArray(int l, int r, vector<long long> &ar){     if (l == r) return getNode(ar[l]);     int mid = (l + r) >> 1;         // Call method to return left Node:    Node left = getMaxSumSubArray(l, mid, ar);         // Call method to return right Node:    Node right = getMaxSumSubArray(mid+1, r, ar);         // Return the merged Node:    return merg(left, right); } // Driver codeint main(){     vector<long long> ar = {-2, -5, 6, -2, -3, 1, 5, -6};    int n = ar.size();    Node ans = getMaxSumSubArray(0, n-1, ar);    cout << "Answer is " << ans._max << "\n";     return 0;}
// Java implementation of the approachimport java.util.*;class GFG{static class Node{         // To store the maximum sum    // for a sub-array    int _max;         // To store the maximum prefix    // sum for a sub-array    int _pre;         // To store the maximum suffix    // sum for a sub-array    int _suf;         // To store the total sum    // for a sub-array    int _sum; };  // Function to create a nodestatic Node getNode(int x){    Node a = new Node();    a._max = x;    a._pre = x;    a._suf = x;    a._sum = x;    return a;} // Function to merge the 2 nodes left and rightstatic Node merg(Node l, Node r){         // Creating node ans    Node ans = new Node();     // Initializing all the variables:    ans._max = ans._pre = ans._suf = ans._sum = 0;         // The max prefix sum of ans Node is maximum of    // a) max prefix sum of left Node    // b) sum of left Node + max prefix sum of right Node    // c) sum of left Node + sum of right Node    ans._pre = Arrays.stream(new int[]{l._pre, l._sum+r._pre,                                       l._sum+r._sum}).max().getAsInt();     // The max suffix sum of ans Node is maximum of    // a) max suffix sum of right Node    // b) sum of right Node + max suffix sum of left Node    // c) sum of left Node + sum of right Node    ans._suf = Arrays.stream(new int[]{r._suf, r._sum+l._suf,                                       l._sum+r._sum}).max().getAsInt();         // Total sum of ans Node = total sum of  // left Node + total sum of right Node    ans._sum = l._sum + r._sum;         // The max sum of ans Node stores    // the answer which is the maximum value among:    // prefix sum of ans Node    // suffix sum of ans Node    // maximum value of left Node    // maximum value of right Node    // prefix value of right Node + suffix value of left Node    ans._max = Arrays.stream(new int[]{ans._pre,                                       ans._suf,                                       ans._sum,                                       l._max, r._max,                                       l._suf+r._pre}).max().getAsInt();     // Return the ans Node    return ans;} // Function for calculating the// max_sum_subArray using divide and conquerstatic Node getMaxSumSubArray(int l, int r, int []ar){     if (l == r) return getNode(ar[l]);    int mid = (l + r) >> 1;         // Call method to return left Node:    Node left = getMaxSumSubArray(l, mid, ar);         // Call method to return right Node:    Node right = getMaxSumSubArray(mid + 1, r, ar);         // Return the merged Node:    return merg(left, right); } // Driver codepublic static void main(String[] args){    int []ar = {-2, -5, 6, -2, -3, 1, 5, -6};    int n = ar.length;    Node ans = getMaxSumSubArray(0, n - 1, ar);    System.out.print("Answer is " +  ans._max + "\n");}} // This code is contributed by shikhasingrajput
# Python3 implementation of the approach class Node:         def __init__(self, x):                 # To store the maximum sum for a sub-array        self._max = x                 # To store the maximum prefix sum for a sub-array        self._pre = x                 # To store the maximum suffix sum for a sub-array        self._suf = x                 # To store the total sum for a sub-array        self._sum = x         # Function to merge the 2 nodes left and rightdef merg(l, r):         # Creating node ans    ans = Node(0)     # The max prefix sum of ans Node is maximum of    # a) max prefix sum of left Node    # b) sum of left Node + max prefix sum of right Node    # c) sum of left Node + sum of right Node    ans._pre = max(l._pre, l._sum+r._pre, l._sum+r._sum)     # The max suffix sum of ans Node is maximum of    # a) max suffix sum of right Node    # b) sum of right Node + max suffix sum of left Node    # c) sum of left Node + sum of right Node    ans._suf = max(r._suf, r._sum+l._suf, l._sum+r._sum)         # Total sum of ans Node = total sum of    # left Node + total sum of right Node    ans._sum = l._sum + r._sum         # The max sum of ans Node stores the answer    # which is the maximum value among:    # prefix sum of ans Node    # suffix sum of ans Node    # maximum value of left Node    # maximum value of right Node    # prefix value of left Node + suffix value of right Node    ans._max = max(ans._pre, ans._suf, ans._sum,                    l._max, r._max, l._suf+r._pre)     # Return the ans Node    return ans # Function for calculating the# max_sum_subArray using divide and conquerdef getMaxSumSubArray(l, r, ar):     if l == r: return Node(ar[l])     mid = (l + r) // 2         # Call method to return left Node:    left = getMaxSumSubArray(l, mid, ar)         # Call method to return right Node:    right = getMaxSumSubArray(mid+1, r, ar)         # Return the merged Node:    return merg(left, right) # Driver codeif __name__ == "__main__":     ar = [-2, -5, 6, -2, -3, 1, 5, -6]    n = len(ar)    ans = getMaxSumSubArray(0, n-1, ar)    print("Answer is", ans._max) # This code is contributed by Rituraj Jain
// C# implementation of the approachusing System;using System.Linq;public class GFG{   class Node{         // To store the maximum sum    // for a sub-array    public int _max;         // To store the maximum prefix    // sum for a sub-array    public int _pre;         // To store the maximum suffix    // sum for a sub-array    public int _suf;         // To store the total sum    // for a sub-array    public int _sum;}; // Function to create a nodestatic Node getNode(int x){    Node a = new Node();    a._max = x;    a._pre = x;    a._suf = x;    a._sum = x;    return a;} // Function to merge the 2 nodes left and rightstatic Node merg(Node l, Node r){         // Creating node ans    Node ans = new Node();     // Initializing all the variables:    ans._max = ans._pre = ans._suf = ans._sum = 0;         // The max prefix sum of ans Node is maximum of    // a) max prefix sum of left Node    // b) sum of left Node + max prefix sum of right Node    // c) sum of left Node + sum of right Node    ans._pre = (new int[]{l._pre, l._sum+r._pre,                                       l._sum+r._sum}).Max();     // The max suffix sum of ans Node is maximum of    // a) max suffix sum of right Node    // b) sum of right Node + max suffix sum of left Node    // c) sum of left Node + sum of right Node    ans._suf = (new int[]{r._suf, r._sum+l._suf,                                       l._sum+r._sum}).Max();         // Total sum of ans Node = total sum of  // left Node + total sum of right Node    ans._sum = l._sum + r._sum;         // The max sum of ans Node stores    // the answer which is the maximum value among:    // prefix sum of ans Node    // suffix sum of ans Node    // maximum value of left Node    // maximum value of right Node    // prefix value of right Node + suffix value of left Node    ans._max = (new int[]{ans._pre,                                       ans._suf,                                       ans._sum,                                       l._max, r._max,                                       l._suf+r._pre}).Max();     // Return the ans Node    return ans;} // Function for calculating the// max_sum_subArray using divide and conquerstatic Node getMaxSumSubArray(int l, int r, int []ar){    if (l == r) return getNode(ar[l]);    int mid = (l + r) >> 1;         // Call method to return left Node:    Node left = getMaxSumSubArray(l, mid, ar);         // Call method to return right Node:    Node right = getMaxSumSubArray(mid + 1, r, ar);         // Return the merged Node:    return merg(left, right);} // Driver codepublic static void Main(String[] args){    int []ar = {-2, -5, 6, -2, -3, 1, 5, -6};    int n = ar.Length;    Node ans = getMaxSumSubArray(0, n - 1, ar);    Console.Write("Answer is " +  ans._max + "\n");}} // This code is contributed by shikhasingrajput
<script> // Javascript implementation of the approachclass Node {     constructor()    {        // To store the maximum sum        // for a sub-array        var _max;                 // To store the maximum prefix        // sum for a sub-array        var _pre;                 // To store the maximum suffix        // sum for a sub-array        var _suf;                 // To store the total sum        // for a sub-array        var _sum;    } };  // Function to create a nodefunction getNode(x){    var a = new Node();    a._max = x;    a._pre = x;    a._suf = x;    a._sum = x;    return a;} // Function to merge the 2 nodes left and rightfunction merg(l, r){         // Creating node ans    var ans = new Node();     // Initializing all the variables:    ans._max = ans._pre = ans._suf = ans._sum = 0;         // The max prefix sum of ans Node is maximum of    // a) max prefix sum of left Node    // b) sum of left Node + max prefix sum of right Node    // c) sum of left Node + sum of right Node    ans._pre = Math.max(l._pre, l._sum+r._pre, l._sum+r._sum);     // The max suffix sum of ans Node is maximum of    // a) max suffix sum of right Node    // b) sum of right Node + max suffix sum of left Node    // c) sum of left Node + sum of right Node    ans._suf = Math.max(r._suf, r._sum+l._suf, l._sum+r._sum);         // Total sum of ans Node = total sum of left Node + total sum of right Node    ans._sum = l._sum + r._sum;         // The max sum of ans Node stores the answer which is the maximum value among:    // prefix sum of ans Node    // suffix sum of ans Node    // maximum value of left Node    // maximum value of right Node    // prefix value of right Node + suffix value of left Node    ans._max = Math.max(ans._pre, ans._suf, ans._sum,l._max, r._max, l._suf+r._pre);     // Return the ans Node    return ans;} // Function for calculating the// max_sum_subArray using divide and conquerfunction getMaxSumSubArray(l, r, ar){     if (l == r) return getNode(ar[l]);     var mid = (l + r) >> 1;         // Call method to return left Node:    var left = getMaxSumSubArray(l, mid, ar);         // Call method to return right Node:    var right = getMaxSumSubArray(mid+1, r, ar);         // Return the merged Node:    return merg(left, right); } // Driver codevar ar = [-2, -5, 6, -2, -3, 1, 5, -6];var n = ar.length;var ans = getMaxSumSubArray(0, n-1, ar);document.write("Answer is " + ans._max + "<br>"); // This code is contributed by rutvik_56.</script>
Output: 
Answer is 7
Time Complexity: The getMaxSumSubArray() recursive function generates the following recurrence relation. T(n) = 2 * T(n / 2) + O(1) note that conquer part takes only O(1) time. So on solving this recurrence using Master’s Theorem we get the time complexity of O(n). 
rituraj_jain
Akanksha_Rai
hitshob469
shikhasingrajput
rutvik_56
subarray
subarray-sum
Arrays
Divide and Conquer
Arrays
Divide and Conquer
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