Print matrix after applying increment operations in M ranges

Given a 2-D matrix mat[][] of size N * N, initially all the elements of the matrix are 0. A number of queries(M ranges) need to be performed on the matrix where each query consists of four integers X1, Y1, X2 and Y2, the task is to add 1 to all the cells between mat[X1][Y1] and mat[X2][Y2] (including both) and print the contents of the updated matrix in the end.

Examples:

Input: N = 2, q[][] = { { 0, 0, 1, 1 }, { 0, 0, 0, 1 } } 
Output:
2 2 
1 1 
After 1st query: mat[][] = { {1, 1}, {1, 1} } 
After 2nd query: mat[][] = { {2, 2}, {1, 1} }

Input: N = 5, q[][] = { { 0, 0, 1, 2 }, { 1, 2, 3, 4 }, { 1, 4, 3, 4 } } 
Output:
1 1 1 1 1 
1 1 2 1 2 
2 2 2 2 2 
2 2 2 2 2 
0 0 0 0 0 

Approach: For each query (X1, Y1) represents the top left cell of the sub-matrix and (X2, Y2) represents the bottom right cell of the sub-matrix. For each top left cell add 1 to the top left element and subtract 1 from the element next to bottom right cell (if any). 

Then maintain a running sum of all the elements from the original (now modified) matrix and at every addition, the current sum is the element (updated) at the current position.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include<bits/stdc++.h>
using namespace std;

// Function to update and print the
// matrix after performing queries
void updateMatrix(int n, int q[3][4])
{
    int i, j;
    int mat[n][n];
    for(int i = 0; i < n; i++)
    for(int j = 0; j < n; j++)
    mat[i][j] = 0;
    for (i = 0; i < 3; i++)
    {
        int X1 = q[i][0];
        int Y1 = q[i][1];
        int X2 = q[i][2];
        int Y2 = q[i][3];

        // Add 1 to the first element of
        // the sub-matrix
        mat[X1][Y1]++;

        // If there is an element after the
        // last element of the sub-matrix
        // then decrement it by 1
        if (Y2 + 1 < n)
            mat[X2][Y2 + 1]--;
        else if (X2 + 1 < n)
            mat[X2 + 1][0]--;
    }

    // Calculate the running sum
    int sum = 0;
    for (i = 0; i < n; i++)
    {
        for (j = 0; j < n; j++)
        {
            sum += mat[i][j];

            // Print the updated element
            cout << sum << " ";
        }

        // Next line
        cout << endl;
    }
}

// Driver code
int main()
{

    // Size of the matrix
    int n = 5;


    // Queries
    int q[3][4] = {{ 0, 0, 1, 2 },
                   { 1, 2, 3, 4 },
                   { 1, 4, 3, 4 }};

    updateMatrix(n, q);
    return 0;
}

// This code is contributed by chandan_jnu

Java

// Java implementation of the approach
public class GFG {

    // Function to update and print the matrix
    // after performing queries
    static void updateMatrix(int n, int q[][], int mat[][])
    {
        int i, j;
        for (i = 0; i < q.length; i++) {
            int X1 = q[i][0];
            int Y1 = q[i][1];
            int X2 = q[i][2];
            int Y2 = q[i][3];

            // Add 1 to the first element of the sub-matrix
            mat[X1][Y1]++;

            // If there is an element after the last element
            // of the sub-matrix then decrement it by 1
            if (Y2 + 1 < n)
                mat[X2][Y2 + 1]--;
            else if (X2 + 1 < n)
                mat[X2 + 1][0]--;
        }

        // Calculate the running sum
        int sum = 0;
        for (i = 0; i < n; i++) {
            for (j = 0; j < n; j++) {
                sum += mat[i][j];

                // Print the updated element
                System.out.print(sum + " ");
            }

            // Next line
            System.out.println();
        }
    }

    // Driver code
    public static void main(String[] args)
    {

        // Size of the matrix
        int n = 5;
        int mat[][] = new int[n][n];

        // Queries
        int q[][] = { { 0, 0, 1, 2 },
                      { 1, 2, 3, 4 },
                      { 1, 4, 3, 4 } };

        updateMatrix(n, q, mat);
    }
}

Python3

# Python 3 implementation of the approach

# Function to update and print the matrix
# after performing queries
def updateMatrix(n, q, mat):

    for i in range(0, len(q)):
            X1 = q[i][0];
            Y1 = q[i][1];
            X2 = q[i][2];
            Y2 = q[i][3];

            # Add 1 to the first element of
            # the sub-matrix
            mat[X1][Y1] = mat[X1][Y1] + 1;

            # If there is an element after the
            # last element of the sub-matrix
            # then decrement it by 1
            if (Y2 + 1 < n):
                mat[X2][Y2 + 1] = mat[X2][Y2 + 1] - 1;
            elif (X2 + 1 < n):
                mat[X2 + 1][0] = mat[X2 + 1][0] - 1;

    # Calculate the running sum
    sum = 0;
    for i in range(0, n):
        for j in range(0, n):
            sum =sum + mat[i][j];

            # Print the updated element
            print(sum, end = ' ');

        # Next line
        print(" ");

# Driver code

# Size of the matrix
n = 5;
mat = [[0 for i in range(n)]
          for i in range(n)];

# Queries
q = [[ 0, 0, 1, 2 ],
     [ 1, 2, 3, 4 ],
     [ 1, 4, 3, 4 ]];

updateMatrix(n, q, mat);

# This code is contributed
# by Shivi_Aggarwal

C#

// C# implementation of the above approach

using System;

public class GFG {

    // Function to update and print the matrix
    // after performing queries
    static void updateMatrix(int n, int [,]q, int [,]mat)
    {
        int i, j;
        for (i = 0; i < q.GetLength(0); i++) {
            int X1 = q[i,0];
            int Y1 = q[i,1];
            int X2 = q[i,2];
            int Y2 = q[i,3];

            // Add 1 to the first element of the sub-matrix
            mat[X1,Y1]++;

            // If there is an element after the last element
            // of the sub-matrix then decrement it by 1
            if (Y2 + 1 < n)
                mat[X2,Y2 + 1]--;
            else if (X2 + 1 < n)
                mat[X2 + 1,0]--;
        }

        // Calculate the running sum
        int sum = 0;
        for (i = 0; i < n; i++) {
            for (j = 0; j < n; j++) {
                sum += mat[i,j];

                // Print the updated element
                Console.Write(sum + " ");
            }

            // Next line
            Console.WriteLine();
        }
    }

    // Driver code
    public static void Main()
    {

        // Size of the matrix
        int n = 5;
        int [,]mat = new int[n,n];

        // Queries
        int [,]q = { { 0, 0, 1, 2 },
                    { 1, 2, 3, 4 },
                    { 1, 4, 3, 4 } };

        updateMatrix(n, q, mat);
    }
    // This code is contributed by Ryuga
}

PHP

<?php
// PHP implementation of the approach

// Function to update and print the
// matrix after performing queries
function updateMatrix($n, $q, $mat)
{
    for ($i = 0; $i < sizeof($q); $i++)
    {
        $X1 = $q[$i][0];
        $Y1 = $q[$i][1];
        $X2 = $q[$i][2];
        $Y2 = $q[$i][3];

        // Add 1 to the first element of
        // the sub-matrix
        $mat[$X1][$Y1]++;

        // If there is an element after the last
        // element of the sub-matrix then decrement
        // it by 1
        if ($Y2 + 1 < $n)
            $mat[$X2][$Y2 + 1]--;
        else if ($X2 + 1 < $n)
            $mat[$X2 + 1][0]--;
    }

    // Calculate the running sum
    $sum = 0;
    for ($i = 0; $i < $n; $i++)
    {
        for ($j = 0; $j < $n; $j++)
        {
            $sum += $mat[$i][$j];

            // Print the updated element
            echo($sum . " ");
        }

        // Next line
        echo("\n");
    }
}

// Driver code

// Size of the matrix
$n = 5;
$mat = array_fill(0, $n,
       array_fill(0, $n, 0));

// Queries
$q = array(array( 0, 0, 1, 2 ),
           array( 1, 2, 3, 4 ),
           array( 1, 4, 3, 4 ));

updateMatrix($n, $q, $mat);

// This code is contributed by chandan_jnu
?>

Javascript

<script>
    // Javascript implementation of the approach

    // Function to update and print the matrix
    // after performing queries
    function updateMatrix(n, q, mat)
    {
        let i, j;
        for (i = 0; i < q.length; i++) {
            let X1 = q[i][0];
            let Y1 = q[i][1];
            let X2 = q[i][2];
            let Y2 = q[i][3];

            // Add 1 to the first element of the sub-matrix
            mat[X1][Y1]++;

            // If there is an element after the last element
            // of the sub-matrix then decrement it by 1
            if (Y2 + 1 < n)
                mat[X2][Y2 + 1]--;
            else if (X2 + 1 < n)
                mat[X2 + 1][0]--;
        }

        // Calculate the running sum
        let sum = 0;
        for (i = 0; i < n; i++) {
            for (j = 0; j < n; j++) {
                sum += mat[i][j];

                // Print the updated element
                document.write(sum + " ");
            }

            // Next line
            document.write("</br>");
        }
    }

    // Size of the matrix
    let n = 5;
    let mat = new Array(n);
    for(let i = 0; i < n; i++)
    {
        mat[i] = new Array(n);
        for(let j = 0; j < n; j++)
        {
            mat[i][j] = 0;
        }
    }

    // Queries
    let q = [ [ 0, 0, 1, 2 ],
              [ 1, 2, 3, 4 ],
              [ 1, 4, 3, 4 ] ];

    updateMatrix(n, q, mat);

    // This code is contributed by divyeshrabadiya07.
</script>

Output

1 1 1 1 1 
1 1 2 1 2 
2 2 2 2 2 
2 2 2 2 2 
0 0 0 0 0 

Complexity Analysis:

• Time Complexity: O(n²)
• Auxiliary Space: O(n²)