Generate a matrix with each row and column of given sum
Given two arrays row[] and col[] of size N and M respectively, the task is to construct a matrix of dimensions N × M such that the sum of matrix elements in every ith row is row[i] and the sum of matrix elements in every jth column is col[j].
Examples:
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Input: row[] = {5, 7, 10}, col[] = {8, 6, 8}
Output: {{0, 5, 0}, {6, 1, 0}, {2, 0, 8}}
Explanation:
Row 1 has sum equal to 5 and column 1 has sum equal to 8
Row 2 has sum equal to 7 and column 2 has sum equal to 6
Row 3 has sum equal to 10 and column 3 has sum equal to 8Input: row[] ={1, 0}, col[] = {1}
Output: {{1}, {0}}
Explanation:
Row 1 has sum equal to 1 and column 1 has sum equal to 1
Row 2 has sum 0
Approach: The simplest approach to solve this problem is to use a Greedy Approach. Follow the steps below to solve this problem:
- Initialize a matrix having dimensions N × M.
- Start filling each cell (i, j) of the matrix in the following manner:
- For each cell (i, j), choose the minimum value of row[i], col[j], and place it at cell (i, j). Let it be minm.
- Subtract minm from row[i] and col[j].
- After completing the above steps, print the matrix formed.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to generate a matrix with// sum of each row equal to sum of r[]// and sum of each column equal to sum of c[]vector<vector<int> > restoreGem( vector<int>& r, vector<int>& c){ // Initialize a matrix vector<vector<int> > dp(r.size(), vector<int>(c.size(), 0)); // Traverse each cell (i, j) of the matrix for (int i = 0; i < r.size(); i++) { for (int j = 0; j < c.size(); j++) { // Assign the minimum of the // row or column value int m = min(r[i], c[j]); dp[i][j] = m; // Subtract the minimum from // both row and column sum r[i] -= m; c[j] -= m; } } return dp;}void printMatrix(vector<vector<int> > ans, int N, int M){ // Print the matrix obtained for (int i = 0; i < N; i++) { for (int j = 0; j < M; j++) { cout << ans[i][j] << " "; } cout << endl; }}// Driver Codeint main(){ vector<int> rowSum = { 5, 7, 10 }; vector<int> colSum = { 8, 6, 8 }; vector<vector<int> > ans = restoreGem(rowSum, colSum); printMatrix(ans, rowSum.size(), colSum.size());} |
Java
// Java program for the above approachimport java.io.*;class GFG{// Function to generate a matrix with// sum of each row equal to sum of r[]// and sum of each column equal to sum of c[]static int[][] restoreGem(int[] r, int[] c){ // Initialize a matrix int[][] dp = new int[r.length][c.length]; // Traverse each cell (i, j) of the matrix for (int i = 0; i < r.length; i++) { for (int j = 0; j < c.length; j++) { // Assign the minimum of the // row or column value int m = Math.min(r[i], c[j]); dp[i][j] = m; // Subtract the minimum from // both row and column sum r[i] -= m; c[j] -= m; } } return dp;}static void printMatrix(int[][] ans, int N, int M){ // Print the matrix obtained for (int i = 0; i < N; i++) { for (int j = 0; j < M; j++) { System.out.print(ans[i][j] + " "); } System.out.println(); }}// Driver Codepublic static void main(String[] args){ int[] rowSum = { 5, 7, 10 }; int[] colSum = { 8, 6, 8 }; int[][] ans = restoreGem(rowSum, colSum); printMatrix(ans, rowSum.length, colSum.length);}}// This code is contributed by susmitakundugoaldanga. |
C#
// C# program for the above approachusing System;public class GFG{// Function to generate a matrix with// sum of each row equal to sum of r[]// and sum of each column equal to sum of c[]static int[,] restoreGem(int[] r, int[] c){ // Initialize a matrix int[,] dp = new int[r.Length, c.Length]; // Traverse each cell (i, j) of the matrix for (int i = 0; i < r.Length; i++) { for (int j = 0; j < c.Length; j++) { // Assign the minimum of the // row or column value int m = Math.Min(r[i], c[j]); dp[i,j] = m; // Subtract the minimum from // both row and column sum r[i] -= m; c[j] -= m; } } return dp;}static void printMatrix(int[,] ans, int N, int M){ // Print the matrix obtained for (int i = 0; i < N; i++) { for (int j = 0; j < M; j++) { Console.Write(ans[i, j] + " "); } Console.WriteLine(); }}// Driver Codepublic static void Main(String[] args){ int[] rowSum = { 5, 7, 10 }; int[] colSum = { 8, 6, 8 }; int[,] ans = restoreGem(rowSum, colSum); printMatrix(ans, rowSum.Length, colSum.Length);}}// This code is contributed by 29AjayKumar |
Javascript
<script>// javascript program of the above approach// Function to generate a matrix with// sum of each row equal to sum of r[]// and sum of each column equal to sum of c[]function restoreGem(r, c){ // Initialize a matrix let dp = new Array(r.length); // Loop to create 2D array using 1D array for (var i = 0; i < dp.length; i++) { dp[i] = new Array(2); } // Traverse each cell (i, j) of the matrix for (let i = 0; i < r.length; i++) { for (let j = 0; j < c.length; j++) { // Assign the minimum of the // row or column value let m = Math.min(r[i], c[j]); dp[i][j] = m; // Subtract the minimum from // both row and column sum r[i] -= m; c[j] -= m; } } return dp;}function prletMatrix(ans, N, M){ // Prlet the matrix obtained for (let i = 0; i < N; i++) { for (let j = 0; j < M; j++) { document.write(ans[i][j] + " "); } document.write("<br/>"); }} // Driver Code let rowSum = [ 5, 7, 10 ]; let colSum = [ 8, 6, 8 ]; let ans = restoreGem(rowSum, colSum); prletMatrix(ans, rowSum.length, colSum.length);</script> |
5 0 0 3 4 0 0 2 8
Time Complexity: O(N×M)
Auxiliary Space: O(N×M)
