Given two 2D arrays b[][] and c[][] of N rows and M columns. The task is to minimise the absolute difference of the sum of b[i][j]s and the sum of c[i][j]s along the path from (0, 0) to (N – 1, M – 1).
Examples:
Input: b[][] = {{1, 4}, {2, 4}}, c[][] = {{3, 2}, {3, 1}}
Output: 0
Choose path (0, 0) -> (1, 0) -> (1, 1)
sum of b[i][j]s are = 1 + 2 + 4 = 7
sum of c[i][j]s are = 3 + 3 + 1 = 7
absolute difference is zero
Input: b[][] = {{1, 10, 50}, {50, 10, 1}}, c[][] = {{1, 2, 3}, {4, 5, 6}}
Output: 2
Approach: The answer is independent from the order of deciding b[i][j] and c[i][j] and the path. So let’s consider a boolean table such that dp[i][j][k] will be true if (i, j) can be reached with minimum difference of k.
If it is true then for the cell (i + 1, j) it is either k + |bi+1, j – ci+1, j| or |k – |bi+1, j – ci+1, j||. The same is true for square (i, j + 1). Therefore, the table can be filled in the increasing order of i and j.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define MAXI 50int dp[MAXI][MAXI][MAXI * MAXI];int n, m;// Function to return the minimum difference// path from (0, 0) to (N - 1, M - 1)int minDifference(int x, int y, int k, vector<vector<int> > b, vector<vector<int> > c){ // Terminating case if (x >= n or y >= m) return INT_MAX; // Base case if (x == n - 1 and y == m - 1) { int diff = b[x][y] - c[x][y]; return min(abs(k - diff), abs(k + diff)); } int& ans = dp[x][y][k]; // If it is already visited if (ans != -1) return ans; ans = INT_MAX; int diff = b[x][y] - c[x][y]; // Recursive calls ans = min(ans, minDifference(x + 1, y, abs(k + diff), b, c)); ans = min(ans, minDifference(x, y + 1, abs(k + diff), b, c)); ans = min(ans, minDifference(x + 1, y, abs(k - diff), b, c)); ans = min(ans, minDifference(x, y + 1, abs(k - diff), b, c)); // Return the value return ans;}// Driver codeint main(){ n = 2, m = 2; vector<vector<int> > b = { { 1, 4 }, { 2, 4 } }; vector<vector<int> > c = { { 3, 2 }, { 3, 1 } }; memset(dp, -1, sizeof(dp)); // Function call cout << minDifference(0, 0, 0, b, c); return 0;} |
Java
// Java implementation of the approach class GFG { final static int MAXI = 50 ; static int dp[][][] = new int[MAXI][MAXI][MAXI * MAXI]; static int n, m; final static int INT_MAX = Integer.MAX_VALUE; // Function to return the minimum difference // path from (0, 0) to (N - 1, M - 1) static int minDifference(int x, int y, int k, int b[][], int c[][]) { // Terminating case if (x >= n || y >= m) return INT_MAX; // Base case if (x == n - 1 && y == m - 1) { int diff = b[x][y] - c[x][y]; return Math.min(Math.abs(k - diff), Math.abs(k + diff)); } int ans = dp[x][y][k]; // If it is already visited if (ans != -1) return ans; ans = INT_MAX; int diff = b[x][y] - c[x][y]; // Recursive calls ans = Math.min(ans, minDifference(x + 1, y, Math.abs(k + diff), b, c)); ans = Math.min(ans, minDifference(x, y + 1, Math.abs(k + diff), b, c)); ans = Math.min(ans, minDifference(x + 1, y, Math.abs(k - diff), b, c)); ans = Math.min(ans, minDifference(x, y + 1, Math.abs(k - diff), b, c)); // Return the value return ans; } // Driver code public static void main (String[] args) { n = 2; m = 2; int b[][] = { { 1, 4 }, { 2, 4 } }; int c[][] = { { 3, 2 }, { 3, 1 } }; for(int i = 0; i < MAXI; i++) { for(int j = 0; j < MAXI; j++) { for(int k = 0; k < MAXI * MAXI; k++) { dp[i][j][k] = -1; } } } // Function call System.out.println(minDifference(0, 0, 0, b, c)); } }// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the approach import numpy as npimport sysMAXI = 50INT_MAX = sys.maxsizedp = np.ones((MAXI, MAXI, MAXI * MAXI));dp *= -1# Function to return the minimum difference # path from (0, 0) to (N - 1, M - 1) def minDifference(x, y, k, b, c) : # Terminating case if (x >= n or y >= m) : return INT_MAX; # Base case if (x == n - 1 and y == m - 1) : diff = b[x][y] - c[x][y]; return min(abs(k - diff), abs(k + diff)); ans = dp[x][y][k]; # If it is already visited if (ans != -1) : return ans; ans = INT_MAX; diff = b[x][y] - c[x][y]; # Recursive calls ans = min(ans, minDifference(x + 1, y, abs(k + diff), b, c)); ans = min(ans, minDifference(x, y + 1, abs(k + diff), b, c)); ans = min(ans, minDifference(x + 1, y, abs(k - diff), b, c)); ans = min(ans, minDifference(x, y + 1, abs(k - diff), b, c)); # Return the value return ans; # Driver code if __name__ == "__main__" : n = 2; m = 2; b = [ [ 1, 4 ], [ 2, 4 ] ]; c = [ [ 3, 2 ], [ 3, 1 ] ]; # Function call print(minDifference(0, 0, 0, b, c)); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approach using System;class GFG { static int MAXI = 50 ; static int [,,]dp = new int[MAXI, MAXI, MAXI * MAXI]; static int n, m; static int INT_MAX = int.MaxValue; // Function to return the minimum difference // path from (0, 0) to (N - 1, M - 1) static int minDifference(int x, int y, int k, int [,]b, int [,]c) { int diff = 0; // Terminating case if (x >= n || y >= m) return INT_MAX; // Base case if (x == n - 1 && y == m - 1) { diff = b[x, y] - c[x, y]; return Math.Min(Math.Abs(k - diff), Math.Abs(k + diff)); } int ans = dp[x, y, k]; // If it is already visited if (ans != -1) return ans; ans = INT_MAX; diff = b[x, y] - c[x, y]; // Recursive calls ans = Math.Min(ans, minDifference(x + 1, y, Math.Abs(k + diff), b, c)); ans = Math.Min(ans, minDifference(x, y + 1, Math.Abs(k + diff), b, c)); ans = Math.Min(ans, minDifference(x + 1, y, Math.Abs(k - diff), b, c)); ans = Math.Min(ans, minDifference(x, y + 1, Math.Abs(k - diff), b, c)); // Return the value return ans; } // Driver code public static void Main () { n = 2; m = 2; int [,]b = { { 1, 4 }, { 2, 4 } }; int [,]c = { { 3, 2 }, { 3, 1 } }; for(int i = 0; i < MAXI; i++) { for(int j = 0; j < MAXI; j++) { for(int k = 0; k < MAXI * MAXI; k++) { dp[i, j, k] = -1; } } } // Function call Console.WriteLine(minDifference(0, 0, 0, b, c)); } } // This code is contributed by AnkitRai01 |
Javascript
<script> // JavaScript implementation of the approach let MAXI = 50 ; let dp = new Array(MAXI); let n, m; let INT_MAX = Number.MAX_VALUE; // Function to return the minimum difference // path from (0, 0) to (N - 1, M - 1) function minDifference(x, y, k, b, c) { // Terminating case if (x >= n || y >= m) return INT_MAX; // Base case if (x == n - 1 && y == m - 1) { let diff = b[x][y] - c[x][y]; return Math.min(Math.abs(k - diff), Math.abs(k + diff)); } let ans = dp[x][y][k]; // If it is already visited if (ans != -1) return ans; ans = INT_MAX; let diff = b[x][y] - c[x][y]; // Recursive calls ans = Math.min(ans, minDifference(x + 1, y, Math.abs(k + diff), b, c)); ans = Math.min(ans, minDifference(x, y + 1, Math.abs(k + diff), b, c)); ans = Math.min(ans, minDifference(x + 1, y, Math.abs(k - diff), b, c)); ans = Math.min(ans, minDifference(x, y + 1, Math.abs(k - diff), b, c)); // Return the value return ans; } n = 2; m = 2; let b = [ [ 1, 4 ], [ 2, 4 ] ]; let c = [ [ 3, 2 ], [ 3, 1 ] ]; for(let i = 0; i < MAXI; i++) { dp[i] = new Array(MAXI); for(let j = 0; j < MAXI; j++) { dp[i][j] = new Array(MAXI * MAXI); for(let k = 0; k < MAXI * MAXI; k++) { dp[i][j][k] = -1; } } } // Function call document.write(minDifference(0, 0, 0, b, c)); </script> |
Output:
0
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