Given a matrix of N rows and M columns. It is given that M is a multiple of 3. The columns are divided into 3 sections, the first section is from 0 to m/3-1, the second section is from m/3 to 2m/3-1 and the third section is from 2m/3 to m. The task is to choose a single element from each row and, in adjacent rows, we cannot select from the same section. We have to maximize the sum of the elements chosen.
Examples:
Input: mat[][] = {
{1, 3, 5, 2, 4, 6},
{6, 4, 5, 1, 3, 2},
{1, 3, 5, 2, 4, 6},
{6, 4, 5, 1, 3, 2},
{6, 4, 5, 1, 3, 2},
{1, 3, 5, 2, 4, 6}}
Output: 35Input: mat[][] = {
{1, 2, 3},
{3, 2, 1},
{5, 4, 2}
Output: 10
Approach: The problem can be solved using dynamic programming solution by storing the subproblems and reusing them. Create a dp[][] array where dp[i][0] represents the sum of elements of rows from 0 to i taking elements from section 1. Similarly, for dp[i][1] and dp[i][2]. So, print the max(dp[n – 1][0], dp[n – 1][1], dp[n – 1][2].
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;const int n = 6, m = 6;// Function to find the maximum valuevoid maxSum(long arr[n][m]){ // Dp table long dp[n + 1][3] = { 0 }; // Fill the dp in bottom // up manner for (int i = 0; i < n; i++) { // Maximum of the three sections long m1 = 0, m2 = 0, m3 = 0; for (int j = 0; j < m; j++) { // Maximum of the first section if ((j / (m / 3)) == 0) { m1 = max(m1, arr[i][j]); } // Maximum of the second section else if ((j / (m / 3)) == 1) { m2 = max(m2, arr[i][j]); } // Maximum of the third section else if ((j / (m / 3)) == 2) { m3 = max(m3, arr[i][j]); } } // If we choose element from section 1, // we cannot have selection from same section // in adjacent rows dp[i + 1][0] = max(dp[i][1], dp[i][2]) + m1; dp[i + 1][1] = max(dp[i][0], dp[i][2]) + m2; dp[i + 1][2] = max(dp[i][1], dp[i][0]) + m3; } // Print the maximum sum cout << max(max(dp[n][0], dp[n][1]), dp[n][2]) << '\n';}// Driver codeint main(){ long arr[n][m] = { { 1, 3, 5, 2, 4, 6 }, { 6, 4, 5, 1, 3, 2 }, { 1, 3, 5, 2, 4, 6 }, { 6, 4, 5, 1, 3, 2 }, { 6, 4, 5, 1, 3, 2 }, { 1, 3, 5, 2, 4, 6 } }; maxSum(arr); return 0;} |
Java
// Java program for the above approachclass GFG{static int n = 6, m = 6;// Function to find the maximum valuestatic void maxSum(long arr[][]){ // Dp table long [][]dp= new long[n + 1][3]; // Fill the dp in bottom // up manner for (int i = 0; i < n; i++) { // Maximum of the three sections long m1 = 0, m2 = 0, m3 = 0; for (int j = 0; j < m; j++) { // Maximum of the first section if ((j / (m / 3)) == 0) { m1 = Math.max(m1, arr[i][j]); } // Maximum of the second section else if ((j / (m / 3)) == 1) { m2 = Math.max(m2, arr[i][j]); } // Maximum of the third section else if ((j / (m / 3)) == 2) { m3 = Math.max(m3, arr[i][j]); } } // If we choose element from section 1, // we cannot have selection from same section // in adjacent rows dp[i + 1][0] = Math.max(dp[i][1], dp[i][2]) + m1; dp[i + 1][1] = Math.max(dp[i][0], dp[i][2]) + m2; dp[i + 1][2] = Math.max(dp[i][1], dp[i][0]) + m3; } // Print the maximum sum System.out.print(Math.max(Math.max(dp[n][0], dp[n][1]), dp[n][2]) + "\n");}// Driver codepublic static void main(String[] args){ long arr[][] = { { 1, 3, 5, 2, 4, 6 }, { 6, 4, 5, 1, 3, 2 }, { 1, 3, 5, 2, 4, 6 }, { 6, 4, 5, 1, 3, 2 }, { 6, 4, 5, 1, 3, 2 }, { 1, 3, 5, 2, 4, 6 } }; maxSum(arr);}}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 program for the above approach import numpy as npn = 6; m = 6; # Function to find the maximum value def maxSum(arr) : # Dp table dp = np.zeros((n + 1, 3)); # Fill the dp in bottom # up manner for i in range(n) : # Maximum of the three sections m1 = 0; m2 = 0; m3 = 0; for j in range(m) : # Maximum of the first section if ((j // (m // 3)) == 0) : m1 = max(m1, arr[i][j]); # Maximum of the second section elif ((j // (m // 3)) == 1) : m2 = max(m2, arr[i][j]); # Maximum of the third section elif ((j // (m // 3)) == 2) : m3 = max(m3, arr[i][j]); # If we choose element from section 1, # we cannot have selection from same section # in adjacent rows dp[i + 1][0] = max(dp[i][1], dp[i][2]) + m1; dp[i + 1][1] = max(dp[i][0], dp[i][2]) + m2; dp[i + 1][2] = max(dp[i][1], dp[i][0]) + m3; # Print the maximum sum print(max(max(dp[n][0], dp[n][1]), dp[n][2])); # Driver code if __name__ == "__main__" : arr = [[ 1, 3, 5, 2, 4, 6 ], [ 6, 4, 5, 1, 3, 2 ], [ 1, 3, 5, 2, 4, 6 ], [ 6, 4, 5, 1, 3, 2 ], [ 6, 4, 5, 1, 3, 2 ], [ 1, 3, 5, 2, 4, 6 ]]; maxSum(arr); # This code is contributed by AnkitRai01 |
C#
// C# program for the above approachusing System;class GFG{static int n = 6, m = 6;// Function to find the maximum valuestatic void maxSum(long [,]arr){ // Dp table long [,]dp = new long[n + 1, 3]; // Fill the dp in bottom // up manner for (int i = 0; i < n; i++) { // Maximum of the three sections long m1 = 0, m2 = 0, m3 = 0; for (int j = 0; j < m; j++) { // Maximum of the first section if ((j / (m / 3)) == 0) { m1 = Math.Max(m1, arr[i, j]); } // Maximum of the second section else if ((j / (m / 3)) == 1) { m2 = Math.Max(m2, arr[i, j]); } // Maximum of the third section else if ((j / (m / 3)) == 2) { m3 = Math.Max(m3, arr[i, j]); } } // If we choose element from section 1, // we cannot have selection from same section // in adjacent rows dp[i + 1, 0] = Math.Max(dp[i, 1], dp[i, 2]) + m1; dp[i + 1, 1] = Math.Max(dp[i, 0], dp[i, 2]) + m2; dp[i + 1, 2] = Math.Max(dp[i, 1], dp[i, 0]) + m3; } // Print the maximum sum Console.Write(Math.Max(Math.Max(dp[n, 0], dp[n, 1]), dp[n, 2]) + "\n");}// Driver codepublic static void Main(String[] args){ long [,]arr = { { 1, 3, 5, 2, 4, 6 }, { 6, 4, 5, 1, 3, 2 }, { 1, 3, 5, 2, 4, 6 }, { 6, 4, 5, 1, 3, 2 }, { 6, 4, 5, 1, 3, 2 }, { 1, 3, 5, 2, 4, 6 } }; maxSum(arr);}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program for the above approachconst n = 6, m = 6;// Function to find the maximum valuefunction maxSum(arr){ // Dp table const dp = new Array(n+1).fill(0).map(() => new Array(3).fill(0)); // Fill the dp in bottom // up manner for (var i = 0; i < n; i++) { // Maximum of the three sections var m1 = 0, m2 = 0, m3 = 0; for (var j = 0; j < m; j++) { // Maximum of the first section if (parseInt(j / (m / 3)) == 0) { m1 = Math.max(m1, arr[i][j]); } // Maximum of the second section else if (parseInt(j / (m / 3)) == 1) { m2 = Math.max(m2, arr[i][j]); } // Maximum of the third section else if (parseInt(j / (m / 3)) == 2) { m3 = Math.max(m3, arr[i][j]); } } // If we choose element from section 1, // we cannot have selection from same section // in adjacent rows dp[i + 1][0] = Math.max(dp[i][1], dp[i][2]) + m1; dp[i + 1][1] = Math.max(dp[i][0], dp[i][2]) + m2; dp[i + 1][2] = Math.max(dp[i][1], dp[i][0]) + m3; } // Print the maximum sum document.write( parseInt(Math.max(Math.max(dp[n][0], dp[n][1]), dp[n][2])) + "<br>");}arr = [ [ 1, 3, 5, 2, 4, 6 ], [ 6, 4, 5, 1, 3, 2 ], [ 1, 3, 5, 2, 4, 6 ], [ 6, 4, 5, 1, 3, 2 ], [ 6, 4, 5, 1, 3, 2 ], [ 1, 3, 5, 2, 4, 6 ] ];maxSum(arr);//This code is contributed by SoumikMondal</script> |
Output
35
Time Complexity: O(n*m), Where n and m are the numbers of rows and number columns in the matrix.
Auxiliary Space: O(n), as we are using extra space for the dp matrix.
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