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Find the minimum difference path from (0, 0) to (N-1, M-1)
  • Difficulty Level : Hard
  • Last Updated : 03 Jan, 2020

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 bollean 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 50
  
int 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 code
int 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 np
import sys
  
MAXI = 50
  
INT_MAX = sys.maxsize
  
dp = 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 
Output:
0



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