给定矩阵中相邻行中相同列元素的所有绝对差的最小和

给定一个具有N行和M列的矩阵mat[][] ，任务是找到两个相邻行之间的最小距离，其中两行之间的距离定义为存在于同一列的两个元素之间的所有绝对差的总和在两排。

例子：

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

方法：给定的问题是一个基于实现的问题，可以通过逐行遍历矩阵并计算所有相邻行对的距离来解决。保持变量中所有计算距离的最小值，这是所需的答案。

以下是上述方法的实现：

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

       // JavaScript code for the above approach
 
       // Function to find minimum distance
       // between two adjacent rows in mat
       function calcDist(N, M, mat)
       {
        
           // Stores the required value
           let ans = Number.MAX_VALUE;
 
           // Loop to traverse all the
           // pair of rows in mat[][]
           for (let i = 0; i < N - 1; i++) {
               // Stores the distance
               let dist = 0;
 
               // Loop to calculate
               // the distance
               for (let 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;
       }
 
       let mat = [[1, 4, 7, 10],
       [2, 5, 8, 11],
       [6, 9, 3, 2]];
       document.write(calcDist(mat.length, mat[0].length, mat));
 
      // This code is contributed by Potta Lokesh
   

输出

4

时间复杂度： O(N*M)
辅助空间： O(1)