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 <bits/stdc++.h>using namespace std;// Function to find minimum distance// between two adjacent rows in matint calcDist(int N, int M, vector<vector<int> > 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 approachint main(){ vector<vector<int> > 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<ArrayList<Integer> > 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<ArrayList<Integer> > mat = new ArrayList<ArrayList<Integer> >(); ArrayList<Integer> temp1 = new ArrayList<Integer>( Arrays.asList(1, 4, 7, 10)); ArrayList<Integer> temp2 = new ArrayList<Integer>( Arrays.asList(2, 5, 8, 11)); ArrayList<Integer> temp3 = new ArrayList<Integer>( 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 matdef 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 ansif __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
<script> // 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 </script> |
4
Time Complexity: O(N*M)
Auxiliary Space: O(1)
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