MO's Algorithm (Query Square Root Decomposition) | Set 1 (Introduction) - GeeksforGeeks

30 Mar, 2022
Let us consider the following problem to understand MO’s Algorithm.We are given an array and a set of query ranges, we are required to find the sum of every query range.
Example: 
Input:  arr[]   = {1, 1, 2, 1, 3, 4, 5, 2, 8};
        query[] = [0, 4], [1, 3] [2, 4]
Output: Sum of arr[] elements in range [0, 4] is 8
        Sum of arr[] elements in range [1, 3] is 4  
        Sum of arr[] elements in range [2, 4] is 6
A Naive Solution is to run a loop from L to R and calculate the sum of elements in given range for every query [L, R]
C++
Java
Python3
C#
Javascript
// C++ Program to compute sum of ranges for different range// queries.#include <bits/stdc++.h>using namespace std; // Structure to represent a query rangestruct Query{    int L, R;}; // Prints sum of all query ranges. m is number of queries// n is the size of the array.void printQuerySums(int a[], int n, Query q[], int m){    // One by one compute sum of all queries    for (int i=0; i<m; i++)    {        // Left and right boundaries of current range        int L = q[i].L, R = q[i].R;         // Compute sum of current query range        int sum = 0;        for (int j=L; j<=R; j++)            sum += a[j];         // Print sum of current query range        cout << "Sum of [" << L << ", "            << R << "] is "  << sum << endl;    }} // Driver programint main(){    int a[] = {1, 1, 2, 1, 3, 4, 5, 2, 8};    int n = sizeof(a)/sizeof(a[0]);    Query q[] = {{0, 4}, {1, 3}, {2, 4}};    int m = sizeof(q)/sizeof(q[0]);    printQuerySums(a, n, q, m);    return 0;}
// Java Program to compute sum of ranges for different range// queries.import java.util.*;  // Class to represent a query rangeclass Query{    int L;    int R;    Query(int L, int R){        this.L = L;        this.R = R;    }} class GFG{    // Prints sum of all query ranges. m is number of queries    // n is the size of the array.    static void printQuerySums(int a[], int n, ArrayList<Query> q, int m)    {        // One by one compute sum of all queries        for (int i=0; i<m; i++)        {            // Left and right boundaries of current range            int L = q.get(i).L, R = q.get(i).R;                 // Compute sum of current query range            int sum = 0;            for (int j=L; j<=R; j++)                sum += a[j];                 // Print sum of current query range            System.out.println("Sum of [" + L +                           ", " + R + "] is "  + sum);        }    }         // Driver program    public static void main(String argv[])    {        int a[] = {1, 1, 2, 1, 3, 4, 5, 2, 8};        int n = a.length;                 ArrayList<Query> q = new ArrayList<Query>();        q.add(new Query(0,4));        q.add(new Query(1,3));        q.add(new Query(2,4));                 int m = q.size();        printQuerySums(a, n, q, m);    }} // This code is contributed by shivanisinghss2110
# Python program to compute sum of ranges for different range queries. # Function that accepts array and list of queries and print sum of each querydef printQuerySum(arr,Q):         for q in Q: # Traverse through each query        L,R = q # Extract left and right indices        s = 0        for i in range(L,R+1): # Compute sum of current query range            s += arr[i]                     print("Sum of",q,"is",s) # Print sum of current query range # Driver scriptarr = [1, 1, 2, 1, 3, 4, 5, 2, 8]Q = [[0, 4], [1, 3], [2, 4]]printQuerySum(arr,Q)#This code is contributed by Shivam Singh
// C# program to compute sum of ranges for// different range queriesusing System;using System.Collections;   // Class to represent a query rangepublic class Query{    public int L;    public int R;         public Query(int L, int R)    {        this.L = L;        this.R = R;    }}  class GFG{     // Prints sum of all query ranges. m//is number of queries n is the size// of the array.static void printQuerySums(int []a, int n,                       ArrayList q, int m){         // One by one compute sum of all queries    for(int i = 0; i < m; i++)    {                 // Left and right boundaries of        // current range        int L = ((Query)q[i]).L,            R = ((Query)q[i]).R;          // Compute sum of current query range        int sum = 0;        for(int j = L; j <= R; j++)            sum += a[j];                     // Print sum of current query range        Console.Write("Sum of [" + L + ", " +                      R + "] is " + sum + "\n");    }}  // Driver codepublic static void Main(string []argv){    int []a = { 1, 1, 2, 1, 3, 4, 5, 2, 8 };    int n = a.Length;         ArrayList q = new ArrayList();    q.Add(new Query(0, 4));    q.Add(new Query(1, 3));    q.Add(new Query(2, 4));          int m = q.Count;         printQuerySums(a, n, q, m);}} // This code is contributed by pratham76
<script>// Javascript Program to compute sum of ranges for different range// queries. // Class to represent a query rangeclass Query{    constructor(L, R)    {        this.L = L;        this.R = R;    }} // Prints sum of all query ranges. m is number of queries    // n is the size of the array.function printQuerySums(a, n, q, m){     // One by one compute sum of all queries        for (let i = 0; i < m; i++)        {            // Left and right boundaries of current range            let L = q[i].L, R = q[i].R;                  // Compute sum of current query range            let sum = 0;            for (let j = L; j <= R; j++)                sum += a[j];                  // Print sum of current query range            document.write("Sum of [" + L +                           ", " + R + "] is "  + sum+"<br>");        }} // Driver programlet a = [1, 1, 2, 1, 3, 4, 5, 2, 8];let n = a.length; let q = [];q.push(new Query(0,4));q.push(new Query(1,3));q.push(new Query(2,4)); let m = q.length;printQuerySums(a, n, q, m); // This code is contributed by avanitrachhadiya2155</script>
Output: 
Sum of [0, 4] is 8
Sum of [1, 3] is 4
Sum of [2, 4] is 6
The time complexity of above solution is O(mn).The idea of MO’s algorithm is to pre-process all queries so that result of one query can be used in next query. Below are steps.
Let a[0...n-1] be input array and q[0..m-1] be array of queries. 
Sort all queries in a way that queries with L values from 0 to √n – 1 are put together, then all queries from √n to 2*√n – 1, and so on. All queries within a block are sorted in increasing order of R values.Process all queries one by one in a way that every query uses sum computed in the previous query.Let ‘sum’ be sum of previous query.Remove extra elements of previous query. For example if previous query is [0, 8] and current query is [3, 9], then we subtract a[0],a[1] and a[2] from sumAdd new elements of current query. In the same example as above, we add a[9] to sum.
Sort all queries in a way that queries with L values from 0 to √n – 1 are put together, then all queries from √n to 2*√n – 1, and so on. All queries within a block are sorted in increasing order of R values.
Process all queries one by one in a way that every query uses sum computed in the previous query.Let ‘sum’ be sum of previous query.Remove extra elements of previous query. For example if previous query is [0, 8] and current query is [3, 9], then we subtract a[0],a[1] and a[2] from sumAdd new elements of current query. In the same example as above, we add a[9] to sum.
Let ‘sum’ be sum of previous query.
Remove extra elements of previous query. For example if previous query is [0, 8] and current query is [3, 9], then we subtract a[0],a[1] and a[2] from sum
Add new elements of current query. In the same example as above, we add a[9] to sum.
The great thing about this algorithm is, in step 2, index variable for R change at most O(n * √n) times throughout the run and same for L changes its value at most O(m * √n) times (See below, after the code, for details). All these bounds are possible only because the queries are sorted first in blocks of √n size.
The preprocessing part takes O(m Log m) time.
Processing all queries takes O(n * √n) + O(m * √n) = O((m+n) * √n) time. 
Below is the implementation of the above idea.
C++
Java
Python3
// Program to compute sum of ranges for different range// queries#include <bits/stdc++.h>using namespace std; // Variable to represent block size. This is made global// so compare() of sort can use it.int block; // Structure to represent a query rangestruct Query{    int L, R;}; // Function used to sort all queries so that all queries// of the same block are arranged together and within a block,// queries are sorted in increasing order of R values.bool compare(Query x, Query y){    // Different blocks, sort by block.    if (x.L/block != y.L/block)        return x.L/block < y.L/block;     // Same block, sort by R value    return x.R < y.R;} // Prints sum of all query ranges. m is number of queries// n is size of array a[].void queryResults(int a[], int n, Query q[], int m){    // Find block size    block = (int)sqrt(n);     // Sort all queries so that queries of same blocks    // are arranged together.    sort(q, q + m, compare);     // Initialize current L, current R and current sum    int currL = 0, currR = 0;    int currSum = 0;     // Traverse through all queries    for (int i=0; i<m; i++)    {        // L and R values of current range        int L = q[i].L, R = q[i].R;         // Remove extra elements of previous range. For        // example if previous range is [0, 3] and current        // range is [2, 5], then a[0] and a[1] are subtracted        while (currL < L)        {            currSum -= a[currL];            currL++;        }         // Add Elements of current Range        while (currL > L)        {            currSum += a[currL-1];            currL--;        }        while (currR <= R)        {            currSum += a[currR];            currR++;        }         // Remove elements of previous range.  For example        // when previous range is [0, 10] and current range        // is [3, 8], then a[9] and a[10] are subtracted        while (currR > R+1)        {            currSum -= a[currR-1];            currR--;        }         // Print sum of current range        cout << "Sum of [" << L << ", " << R             << "] is "  << currSum << endl;    }} // Driver programint main(){    int a[] = {1, 1, 2, 1, 3, 4, 5, 2, 8};    int n = sizeof(a)/sizeof(a[0]);    Query q[] = {{0, 4}, {1, 3}, {2, 4}};    int m = sizeof(q)/sizeof(q[0]);    queryResults(a, n, q, m);    return 0;}
// Java Program to compute sum of ranges for// different range queries import java.util.*; // Class to represent a query rangeclass Query{    int L;    int R;    Query(int L, int R){        this.L = L;        this.R = R;    }} class MO{     // Prints sum of all query ranges. m is number of queries    // n is size of array a[].    static void queryResults(int a[], int n, ArrayList<Query> q, int m){                 // Find block size        int block = (int) Math.sqrt(n);             // Sort all queries so that queries of same blocks        // are arranged together.        Collections.sort(q, new Comparator<Query>(){                         // Function used to sort all queries so that all queries             // of the same block are arranged together and within a block,            // queries are sorted in increasing order of R values.            public int compare(Query x, Query y){                 // Different blocks, sort by block.                if (x.L/block != y.L/block)                    return (x.L < y.L ? -1 : 1);                 // Same block, sort by R value                return (x.R < y.R ? -1 : 1);            }        });         // Initialize current L, current R and current sum        int currL = 0, currR = 0;        int currSum = 0;             // Traverse through all queries        for (int i=0; i<m; i++)        {            // L and R values of current range            int L = q.get(i).L, R = q.get(i).R;             // Remove extra elements of previous range. For            // example if previous range is [0, 3] and current            // range is [2, 5], then a[0] and a[1] are subtracted            while (currL < L)            {                currSum -= a[currL];                currL++;            }             // Add Elements of current Range            while (currL > L)            {                currSum += a[currL-1];                currL--;            }            while (currR <= R)            {                currSum += a[currR];                currR++;            }             // Remove elements of previous range.  For example            // when previous range is [0, 10] and current range            // is [3, 8], then a[9] and a[10] are subtracted            while (currR > R+1)            {                currSum -= a[currR-1];                currR--;            }             // Print sum of current range            System.out.println("Sum of [" + L +                           ", " + R + "] is "  + currSum);        }    }     // Driver program    public static void main(String argv[]){        ArrayList<Query> q = new ArrayList<Query>();        q.add(new Query(0,4));        q.add(new Query(1,3));        q.add(new Query(2,4));         int a[] = {1, 1, 2, 1, 3, 4, 5, 2, 8};        queryResults(a, a.length, q, q.size());    }}// This code is contributed by Ajay
# Python program to compute sum of ranges for different range queries import math # Function that accepts array and list of queries and print sum of each querydef queryResults(arr,Q):         #Q.sort(): # Sort by L    #sort all queries so that all queries in the increasing order of R values .     Q.sort(key=lambda x: x[1])         # Initialize current L, current R and current sum    currL,currR,currSum = 0,0,0         # Traverse through all queries    for i in range(len(Q)):        L,R = Q[i] # L and R values of current range                 # Remove extra elements from previous range        # if previous range is [0, 3] and current         # range is [2, 5], then a[0] and a[1] are subtracted         while currL<L:            currSum-=arr[currL]            currL+=1                     # Add elements of current range        while currL>L:            currSum+=arr[currL-1]            currL-=1        while currR<=R:            currSum+=arr[currR]            currR+=1                     # Remove elements of previous range        # when previous range is [0, 10] and current range         # is [3, 8], then a[9] and a[10] are subtracted         while currR>R+1:            currSum-=arr[currR-1]            currR-=1                 # Print the sum of current range        print("Sum of",Q[i],"is",currSum) arr = [1, 1, 2, 1, 3, 4, 5, 2, 8]Q = [[0, 4], [1, 3], [2, 4]]queryResults(arr,Q)#This code is contributed by Shivam Singh
Output: 
Sum of [1, 3] is 4
Sum of [0, 4] is 8
Sum of [2, 4] is 6
The output of above program doesn’t print results of queries in same order as input, because queries are sorted. The program can be easily extended to keep the same order.
Important Observations: 
All queries are known beforehead so that they can be preprocessedIt cannot work for problems where we have update operations also mixed with sum queries.MO’s algorithm can only be used for query problems where a query can be computed from results of the previous query. One more such example is maximum or minimum.
All queries are known beforehead so that they can be preprocessed
It cannot work for problems where we have update operations also mixed with sum queries.
MO’s algorithm can only be used for query problems where a query can be computed from results of the previous query. One more such example is maximum or minimum.
Time Complexity Analysis: The function mainly runs a for loop for all sorted queries. Inside the for loop, there are four while queries that move ‘currL’ and ‘currR’. 
How much currR is moved? For each block, queries are sorted in increasing order of R. So, for a block, currR moves in increasing order. In worst case, before beginning of every block, currR at extreme right and current block moves it back the extreme left. This means that for every block, currR moves at most O(n). Since there are O(√n) blocks, total movement of currR is O(n * √n). 
How much currL is moved? Since all queries are sorted in a way that L values are grouped by blocks, movement is O(√n) when we move from one query to another quert. For m queries, total movement of currL is O(m * √n)Please note that a Simple and more Efficient solution to solve this problem is to compute prefix sum for all elements from 0 to n-1. Let the prefix sum be stored in an array preSum[] (The value of preSum[i] stores sum of arr[0..i]). Once we have built preSum[], we can traverse through all queries one by one. For every query [L, R], we return value of preSum[R] – preSum[L]. Here processing every query takes O(1) time. 
Auxiliary Space: O(1)
The idea of this article is to introduce MO’s algorithm with a very simple example. We will soon be discussing more interesting problems using MO’s algorithm.
Range Minimum Query (Square Root Decomposition and Sparse Table)
References: http://blog.anudeep2011.com/mos-algorithm/This article is contributed by Ruchir Garg. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above 
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