Print matrix after applying increment operations in M ranges

• Difficulty Level : Medium
• Last Updated : 13 Sep, 2022

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``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

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Javascript

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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)

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