Open In App

Final Matrix after incrementing submatrices by K in range given by Q queries

Last Updated : 22 Jun, 2021
Improve
Improve
Like Article
Like
Save
Share
Report

Given a 2D matrix mat[][] of size N*M and Q queries of the form {x1, y1, x2, y2, K}. For each query, the task is to add the value K to submatrix from cell (x1, y1) to (x2, y2). Print the matrix after all the queries performed.

Examples:

Input: N = 3, M = 4, mat[][] = {{1, 0, 1, 2}, {0, 2, 4, 1}, {1, 2, 1, 0}}, Q = 1, Queries[][] = {{0, 0, 1, 1, 2}} 
Output: 
3 2 1 2 
2 4 4 1 
1 2 1 0 
Explanation: 
There is only one query i.e., updating the submatrix from cell mat[0][0] to mat[1][1] by increment of 2, the matrix becomes: 
3 2 1 2 
2 4 4 1 
1 2 1 0 

Input: N = 2, M = 3, mat[][] = {{3, 2, 1}, {2, 4, 4}}, Q = 1, Queries[][] = { {0, 1, 1, 2, -1}, {0, 0, 1, 1, 5}} 
Output: 
8 6 0 
7 8 3 
Explanation: 
For query 1, i.e., updating the submatrix from cell mat[0][1] to mat[1][2] by increment of (-1), the matrix becomes: 
3 1 0 
2 3 3 
For query 2, i.e., updating the submatrix from cell mat[0][0] to mat[2][2] by increment of 5, the matrix becomes: 
8 6 0 
7 8 3

Naive Approach: The simplest approach is to iterate over the submatrix and add K to all elements from mat[x1][y1] to mat[x2][y2] for each query. Print the matrix after the above operations.

Time Complexity: O(N*M*Q) 
Auxiliary Space: O(1) 

Efficient Approach: The idea is to use an auxiliary matrix to perform the update operations on the corners of the submatrix cells and then find the prefix sum of the matrix to get the resultant matrix. Below are the steps:

  1. Initialize all elements of the auxiliary matrix say aux[][] to 0.
  2. For each query {x1, y1, x2, y2, K} update the auxiliary matrix as: 
    • aux[x1][y1] += K
    • if(x2 + 1 < N) then aux[x2 + 1][y1] -= K
    • if(x2 + 1 < N && y2 + 1 < N) then aux[x2 + 1][y2 + 1] += K
    • if(y2 + 1 < N) then aux[x1][y2 + 1] -= K
  3. Find the prefix sum of each row of the auxiliary matrix.
  4. Find the prefix sum of each column of the auxiliary matrix.
  5. Now, update the auxiliary matrix as the sum of elements at each respective cell of the auxiliary and the given matrix.
  6. Print the auxiliary matrix after all the above operations.

Below is the illustration for how auxiliary matrix is created and updated for query[][] = {{0, 0, 1, 1, 2}, {0, 1, 2, 3, -1}}:

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
#define N 3
#define M 4
 
// Query data type
struct query {
    int x1, x2, y1, y2, K;
};
 
// Function to update the given query
void updateQuery(int from_x, int from_y,
                 int to_x, int to_y,
                 int k, int aux[][M])
{
    // Update top cell
    aux[from_x][from_y] += k;
 
    // Update bottom left cell
    if (to_x + 1 < N)
        aux[to_x + 1][from_y] -= k;
 
    // Update bottom right cell
    if (to_x + 1 < N && to_y + 1 < M)
        aux[to_x + 1][to_y + 1] += k;
 
    // Update top right cell
    if (to_y + 1 < M)
        aux[from_x][to_y + 1] -= k;
}
 
// Function that updates the matrix
// mat[][] by adding elements of aux[][]
void updateMatrix(int mat[][M], int aux[][M])
{
 
    // Compute the prefix sum of all columns
    for (int i = 0; i < N; i++) {
        for (int j = 1; j < M; j++) {
            aux[i][j] += aux[i][j - 1];
        }
    }
 
    // Compute the prefix sum of all rows
    for (int i = 0; i < M; i++) {
        for (int j = 1; j < N; j++) {
            aux[j][i] += aux[j - 1][i];
        }
    }
 
    // Get the final matrix by adding
    // mat and aux matrix at each cell
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
 
            mat[i][j] += aux[i][j];
        }
    }
}
 
// Function that prints matrix mat[]
void printMatrix(int mat[][M])
{
    // Traverse each row
    for (int i = 0; i < N; i++) {
 
        // Traverse each columns
        for (int j = 0; j < M; j++) {
 
            cout << mat[i][j] << " ";
        }
        cout << "\n";
    }
}
 
// Function that performs each query in
// the given matrix and print the updated
// matrix after each operation performed
void matrixQuery(int mat[][M], int Q,
 query q[])
{
 
    // Initialize all elements to 0
    int aux[N][M] = {};
 
    // Update auxiliary matrix
    // by traversing each query
    for (int i = 0; i < Q; i++) {
 
        // Update Query
        updateQuery(q[i].x1, q[i].x2,
                    q[i].y1, q[i].y2,
                    q[i].K, aux);
    }
 
    // Compute the final answer
    updateMatrix(mat, aux);
 
    // Print the updated matrix
    printMatrix(mat);
}
 
// Driver Code
int main()
{
    // Given Matrix
    int mat[N][M] = { { 1, 0, 1, 2 },
                      { 0, 2, 4, 1 },
                      { 1, 2, 1, 0 } };
 
    int Q = 1;
 
    // Given Queries
    query q[] = { { 0, 0, 1, 1, 2 } };
 
    // Function Call
    matrixQuery(mat, Q, q);
    return 0;
}


Java




// Java program for the above approach
class GFG{
static final int N = 3;
static final int M = 4;
 
// Query data type
static class query
{
    int x1, x2, y1, y2, K;
    public query(int x1, int x2,
                 int y1, int y2, int k)
    {
        this.x1 = x1;
        this.x2 = x2;
        this.y1 = y1;
        this.y2 = y2;
        K = k;
    }
};
 
// Function to update the given query
static void updateQuery(int from_x, int from_y,
                        int to_x, int to_y,
                        int k, int aux[][])
{
    // Update top cell
    aux[from_x][from_y] += k;
 
    // Update bottom left cell
    if (to_x + 1 < N)
        aux[to_x + 1][from_y] -= k;
 
    // Update bottom right cell
    if (to_x + 1 < N && to_y + 1 < M)
        aux[to_x + 1][to_y + 1] += k;
 
    // Update top right cell
    if (to_y + 1 < M)
        aux[from_x][to_y + 1] -= k;
}
 
// Function that updates the matrix
// mat[][] by adding elements of aux[][]
static void updateMatrix(int mat[][],
                         int aux[][])
{
 
    // Compute the prefix sum of all columns
    for (int i = 0; i < N; i++)
    {
        for (int j = 1; j < M; j++)
        {
            aux[i][j] += aux[i][j - 1];
        }
    }
 
    // Compute the prefix sum of all rows
    for (int i = 0; i < M; i++)
    {
        for (int j = 1; j < N; j++)
        {
            aux[j][i] += aux[j - 1][i];
        }
    }
 
    // Get the final matrix by adding
    // mat and aux matrix at each cell
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < M; j++)
        {
            mat[i][j] += aux[i][j];
        }
    }
}
 
// Function that prints matrix mat[]
static void printMatrix(int mat[][])
{
    // Traverse each row
    for (int i = 0; i < N; i++)
    {
        // Traverse each columns
        for (int j = 0; j < M; j++)
        {
            System.out.print(mat[i][j] + " ");
        }
        System.out.print("\n");
    }
}
 
// Function that performs each query in
// the given matrix and print the updated
// matrix after each operation performed
static void matrixQuery(int mat[][],
                        int Q, query q[])
{
    // Initialize all elements to 0
    int [][]aux = new int[N][M];
 
    // Update auxiliary matrix
    // by traversing each query
    for (int i = 0; i < Q; i++)
    {
        // Update Query
        updateQuery(q[i].x1, q[i].x2,
                    q[i].y1, q[i].y2,
                    q[i].K, aux);
    }
 
    // Compute the final answer
    updateMatrix(mat, aux);
 
    // Print the updated matrix
    printMatrix(mat);
}
 
// Driver Code
public static void main(String[] args)
{
    // Given Matrix
    int mat[][] = {{1, 0, 1, 2},
                   {0, 2, 4, 1},
                   {1, 2, 1, 0}};
 
    int Q = 1;
 
    // Given Queries
    query q[] = {new query(0, 0, 1, 1, 2 )};
 
    // Function Call
    matrixQuery(mat, Q, q);
}
}
 
// This code is contributed by 29AjayKumar


Python3




# Python3 program for the above approach
 
# Query data type
 
# Function to update the given query
def updateQuery(from_x, from_y,
                to_x, to_y, k, aux):
     
    # Update top cell
    aux[from_x][from_y] += k
 
    # Update bottom left cell
    if (to_x + 1 < 3):
        aux[to_x + 1][from_y] -= k
 
    # Update bottom right cell
    if (to_x + 1 < 3 and to_y + 1 < 4):
        aux[to_x + 1][to_y + 1] += k
 
    # Update top right cell
    if (to_y + 1 < 4):
        aux[from_x][to_y + 1] -= k
         
    # return aux
 
# Function that updates the matrix
# mat[][] by adding elements of aux[][]
def updatematrix(mat, aux):
 
    # Compute the prefix sum of all columns
    for i in range(3):
        for j in range(1, 4):
            aux[i][j] += aux[i][j - 1]
 
    # Compute the prefix sum of all rows
    for i in range(4):
        for j in range(1, 3):
            aux[j][i] += aux[j - 1][i]
 
    # Get the final matrix by adding
    # mat and aux matrix at each cell
    for i in range(3):
        for j in range(4):
            mat[i][j] += aux[i][j]
             
    # return mat
 
# Function that prints matrix mat[]
def printmatrix(mat):
     
    # Traverse each row
    for i in range(3):
 
        # Traverse each columns
        for j in range(4):
 
            print(mat[i][j], end = " ")
             
        print()
 
# Function that performs each query in
# the given matrix and print the updated
# matrix after each operation performed
def matrixQuery(mat, Q, q):
 
    # Initialize all elements to 0
    aux = [[ 0 for i in range(4)]
               for i in range(3)]
 
    # Update auxiliary matrix
    # by traversing each query
    for i in range(Q):
         
        # Update Query
        updateQuery(q[i][0], q[i][1],
                    q[i][2], q[i][3],
                    q[i][4], aux)
 
    # Compute the final answer
    updatematrix(mat, aux)
 
    # Print the updated matrix
    printmatrix(mat)
 
# Driver Code
if __name__ == '__main__':
     
    # Given 4atrix
    mat = [ [ 1, 0, 1, 2 ],
            [ 0, 2, 4, 1 ],
            [ 1, 2, 1, 0 ] ]
 
    Q = 1
 
    # Given Queries
    q = [ [0, 0, 1, 1, 2 ] ]
 
    # Function Call
    matrixQuery(mat, Q, q)
 
# This code is contributed by mohit kumar 29


C#




// C# program for the above approach
using System;
 
class GFG{
     
static readonly int N = 3;
static readonly int M = 4;
 
// Query data type
class query
{
    public int x1, x2, y1, y2, K;
    public query(int x1, int x2,
                 int y1, int y2,
                 int k)
    {
        this.x1 = x1;
        this.x2 = x2;
        this.y1 = y1;
        this.y2 = y2;
        K = k;
    }
};
 
// Function to update the given query
static void updateQuery(int from_x, int from_y,
                        int to_x, int to_y,
                        int k, int [,]aux)
{
     
    // Update top cell
    aux[from_x, from_y] += k;
 
    // Update bottom left cell
    if (to_x + 1 < N)
        aux[to_x + 1, from_y] -= k;
 
    // Update bottom right cell
    if (to_x + 1 < N && to_y + 1 < M)
        aux[to_x + 1, to_y + 1] += k;
 
    // Update top right cell
    if (to_y + 1 < M)
        aux[from_x, to_y + 1] -= k;
}
 
// Function that updates the matrix
// [,]mat by adding elements of aux[,]
static void updateMatrix(int [,]mat,
                         int [,]aux)
{
 
    // Compute the prefix sum of all columns
    for(int i = 0; i < N; i++)
    {
        for(int j = 1; j < M; j++)
        {
            aux[i, j] += aux[i, j - 1];
        }
    }
 
    // Compute the prefix sum of all rows
    for(int i = 0; i < M; i++)
    {
        for(int j = 1; j < N; j++)
        {
            aux[j, i] += aux[j - 1, i];
        }
    }
 
    // Get the readonly matrix by adding
    // mat and aux matrix at each cell
    for(int i = 0; i < N; i++)
    {
        for(int j = 0; j < M; j++)
        {
            mat[i, j] += aux[i, j];
        }
    }
}
 
// Function that prints matrix []mat
static void printMatrix(int [,]mat)
{
     
    // Traverse each row
    for(int i = 0; i < N; i++)
    {
         
        // Traverse each columns
        for(int j = 0; j < M; j++)
        {
            Console.Write(mat[i, j] + " ");
        }
        Console.Write("\n");
    }
}
 
// Function that performs each query in
// the given matrix and print the updated
// matrix after each operation performed
static void matrixQuery(int [,]mat,
                        int Q, query []q)
{
     
    // Initialize all elements to 0
    int [,]aux = new int[N, M];
 
    // Update auxiliary matrix
    // by traversing each query
    for(int i = 0; i < Q; i++)
    {
        // Update Query
        updateQuery(q[i].x1, q[i].x2,
                    q[i].y1, q[i].y2,
                    q[i].K, aux);
    }
 
    // Compute the readonly answer
    updateMatrix(mat, aux);
 
    // Print the updated matrix
    printMatrix(mat);
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given Matrix
    int [,]mat = { { 1, 0, 1, 2 },
                   { 0, 2, 4, 1 },
                   { 1, 2, 1, 0 } };
 
    int Q = 1;
 
    // Given Queries
    query []q = {new query( 0, 0, 1, 1, 2 )};
 
    // Function call
    matrixQuery(mat, Q, q);
}
}
 
// This code is contributed by 29AjayKumar


Javascript




<script>
// Javascript program for the above approach
 
let N = 3;
let M = 4;
 
// Query data type
class query
{
    constructor(x1,x2,y1,y2,k)
    {
        this.x1 = x1;
        this.x2 = x2;
        this.y1 = y1;
        this.y2 = y2;
        this.K = k;
    }
}
 
// Function to update the given query
function updateQuery(from_x,from_y,to_x,to_y,k,aux)
{
    // Update top cell
    aux[from_x][from_y] += k;
  
    // Update bottom left cell
    if (to_x + 1 < N)
        aux[to_x + 1][from_y] -= k;
  
    // Update bottom right cell
    if (to_x + 1 < N && to_y + 1 < M)
        aux[to_x + 1][to_y + 1] += k;
  
    // Update top right cell
    if (to_y + 1 < M)
        aux[from_x][to_y + 1] -= k;
}
 
// Function that updates the matrix
// mat[][] by adding elements of aux[][]
function updateMatrix(mat,aux)
{
    // Compute the prefix sum of all columns
    for (let i = 0; i < N; i++)
    {
        for (let j = 1; j < M; j++)
        {
            aux[i][j] += aux[i][j - 1];
        }
    }
  
    // Compute the prefix sum of all rows
    for (let i = 0; i < M; i++)
    {
        for (let j = 1; j < N; j++)
        {
            aux[j][i] += aux[j - 1][i];
        }
    }
  
    // Get the final matrix by adding
    // mat and aux matrix at each cell
    for (let i = 0; i < N; i++)
    {
        for (let j = 0; j < M; j++)
        {
            mat[i][j] += aux[i][j];
        }
    }
}
 
// Function that prints matrix mat[]
function printMatrix(mat)
{
    // Traverse each row
    for (let i = 0; i < N; i++)
    {
        // Traverse each columns
        for (let j = 0; j < M; j++)
        {
            document.write(mat[i][j] + " ");
        }
        document.write("<br>");
    }
}
 
// Function that performs each query in
// the given matrix and print the updated
// matrix after each operation performed
function matrixQuery(mat,Q,q)
{
    // Initialize all elements to 0
    let aux = new Array(N);
    for(let i=0;i<N;i++)
    {
        aux[i]=new Array(M);
        for(let j=0;j<M;j++)
        {
            aux[i][j]=0;
        }
    }
  
    // Update auxiliary matrix
    // by traversing each query
    for (let i = 0; i < Q; i++)
    {
        // Update Query
        updateQuery(q[i].x1, q[i].x2,
                    q[i].y1, q[i].y2,
                    q[i].K, aux);
    }
  
    // Compute the final answer
    updateMatrix(mat, aux);
  
    // Print the updated matrix
    printMatrix(mat);
}
 
// Driver Code
 
// Given Matrix
let mat = [[1, 0, 1, 2],
[0, 2, 4, 1],
[1, 2, 1, 0]];
 
let Q = 1;
 
// Given Queries
let q = [new query(0, 0, 1, 1, 2 )];
 
// Function Call
matrixQuery(mat, Q, q);
 
 
// This code is contributed by unknown2108
</script>


Output: 

3 2 1 2 
2 4 4 1 
1 2 1 0

Time Complexity: O(Q + N*M) 
Auxiliary Space: O(N*M)

 



Like Article
Suggest improvement
Previous
Next
Share your thoughts in the comments

Similar Reads