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# Minimum sum of all absolute differences of same column elements in adjacent rows in a given Matrix

Given a matrix mat[][] having N rows and M columns, the task is to find the minimum distance between two adjacent rows where the distance between two rows is defined as the sum of all absolute differences between two elements present at the same column in the two rows.

Examples:

Input: mat[][] = {{1, 4, 7, 10}, {2, 5, 8, 11}, {6, 9, 3, 12}}
Output: 4
Explanation: The distance between the first two rows can be calculated as (2-1) + (5-4) + (8-7) + (11-10) = 4. Similarly, the distance between the 2nd and 3rd row can be calculated as (6-2) + (9-5) + (8-3) + (12-11) = 14. Hence, the minimum distance among all adjacent rows is 4.

Input: mat[][] = {{1, 25, 81}, {2, 36, 100}, {9, 49, 50}, {16, 64, 25}}
Output : 31

Approach:  The given problem is an implementation-based problem that can be solved by iterating through the matrix row-wise and calculating the distances over all pairs of adjacent rows. Maintain the minimum of all the calculated distances in a variable which is the required answer.

Below is the implementation of the above approach :

## C++

 // C++ program of the above approach. #include using namespace std;   // Function to find minimum distance // between two adjacent rows in mat int calcDist(int N, int M, vector > mat) {     // Stores the required value     int ans = INT_MAX;       // Loop to traverse all the     // pair of rows in mat[][]     for (int i = 0; i < N - 1; i++) {         // Stores the distance         int dist = 0;           // Loop to calculate         // the distance         for (int j = 0; j < M; j++) {             dist += abs(mat[i][j] - mat[i + 1][j]);         }           // Update ans         ans = min(ans, dist);     }       // Return Answer     return ans; }   // C++ program of the above approach int main() {     vector > mat = { { 1, 4, 7, 10 },                                  { 2, 5, 8, 11 },                                  { 6, 9, 3, 2 } };     cout << calcDist(mat.size(), mat[0].size(), mat);       return 0; }

## Java

 // JAVA program of the above approach. import java.util.*; class GFG {     // Function to find minimum distance   // between two adjacent rows in mat   public static int     calcDist(int N, int M,              ArrayList > mat)   {       // Stores the required value     int ans = Integer.MAX_VALUE;       // Loop to traverse all the     // pair of rows in mat[][]     for (int i = 0; i < N - 1; i++)     {         // Stores the distance       int dist = 0;         // Loop to calculate       // the distance       for (int j = 0; j < M; j++) {         dist += Math.abs(mat.get(i).get(j)                          - mat.get(i + 1).get(j));       }         // Update ans       ans = Math.min(ans, dist);     }       // Return Answer     return ans;   }     // JAVA program of the above approach   public static void main(String[] args)   {     ArrayList > mat       = new ArrayList >();     ArrayList temp1 = new ArrayList(       Arrays.asList(1, 4, 7, 10));     ArrayList temp2 = new ArrayList(       Arrays.asList(2, 5, 8, 11));     ArrayList temp3 = new ArrayList(       Arrays.asList(6, 9, 3, 2));     mat.add(temp1);     mat.add(temp2);     mat.add(temp3);       System.out.print(       calcDist(mat.size(), mat.get(0).size(), mat));   } }   // This code is contributed by Taranpreet

## Python3

 # python3 program of the above approach. INT_MAX = 2147483647   # Function to find minimum distance # between two adjacent rows in mat def calcDist(N, M, mat):       # Stores the required value     ans = INT_MAX       # Loop to traverse all the     # pair of rows in mat[][]     for i in range(0, N - 1):                 # Stores the distance         dist = 0           # Loop to calculate         # the distance         for j in range(0, M):             dist += abs(mat[i][j] - mat[i + 1][j])           # Update ans         ans = min(ans, dist)       # Return Answer     return ans   if __name__ == "__main__":       mat = [[1, 4, 7, 10],            [2, 5, 8, 11],            [6, 9, 3, 2]]       print(calcDist(len(mat), len(mat[0]), mat))       # This code is contributed by rakeshsahni

## C#

 // C# program of the above approach. using System; class GFG {     // Function to find minimum distance   // between two adjacent rows in mat   static int calcDist(int N, int M, int[, ] mat)   {       // Stores the required value     int ans = Int32.MaxValue;       // Loop to traverse all the     // pair of rows in mat[][]     for (int i = 0; i < N - 1; i++) {       // Stores the distance       int dist = 0;         // Loop to calculate       // the distance       for (int j = 0; j < M; j++) {         dist += Math.Abs(mat[i, j] - mat[i + 1, j]);       }         // Update ans       ans = Math.Min(ans, dist);     }       // Return Answer     return ans;   }     // Criver code   public static void Main()   {     int[, ] mat = { { 1, 4, 7, 10 },                    { 2, 5, 8, 11 },                    { 6, 9, 3, 2 } };     Console.Write(calcDist(mat.GetLength(0),                            mat.GetLength(1), mat));   } }   // This code is contributed by Samim Hossain Mondal.

## Javascript



Output

4

Time Complexity: O(N*M)
Auxiliary Space: O(1)

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