# Find the minimum difference path from (0, 0) to (N-1, M-1)

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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 > b, ` `                  ``vector > 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 > b = { { 1, 4 }, { 2, 4 } }; ` ` `  `    ``vector > 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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