Minimum difference between adjacent elements of array which contain elements from each row of a matrix

3.2

Given a matrix of N rows and M columns, the task is to find the minimum absolute difference between any of the two adjacent elements of an array of size N, which is created by picking one element from each row of the matrix. Note the element picked from row 1 will become arr[0], element picked from row 2 will become arr[1] and so on.

Examples:

Input :  N = 2, M = 2
m[2][2] = { 8, 2,
            6, 8 }
Output : 0.
Picking 8 from row 1 and picking 8 from row 2, we create an array { 8, 8 } and minimum
difference between any of adjacent element is 0.

Input :  N = 3, M = 3
m[3][3] = { 1, 2, 3
            4, 5, 6 
            7, 8, 9 }
Output : 1.

The idea is to sort all rows individually and then do binary search to find the closest element in next row for each element.
To do this in an efficient manner, sort each row of the matrix. Starting from row 1 to row N – 1 of matrix, for each element m[i][j] of current row in the matrix, find the smallest element in the next row which is greater than or equal to the current element, say p and the largest element which is smaller than the current element, say q. This can be done using Binary Search. Finally,find the minimum of the difference of current element from p and q and update the variable.

Below is implementation of this approach:

C/C++

// C++ program to find the minimum absolute difference
// between any of the adjacent elements of an array
// which is created by picking one element from each
// row of the matrix.
#include<bits/stdc++.h>
using namespace std;
#define R 2
#define C 2

// Return smallest element greater than or equal
// to the current element.
int bsearch(int low, int high, int n, int arr[])
{
    int mid = (low + high)/2;

    if(low <= high)
    {
        if(arr[mid] < n)
            return bsearch(mid +1, high, n, arr);
        return bsearch(low, mid - 1, n, arr);
    }

    return low;
}

// Return the minimum absolute difference adjacent
// elements of array
int mindiff(int arr[R][C], int n, int m)
{
    // Sort each row of the matrix.
    for (int i = 0; i < n; i++)
        sort(arr[i], arr[i] + m);

    int ans = INT_MAX;

    // For each matrix element
    for (int i = 0; i < n - 1; i++)
    {
        for (int j = 0; j < m; j++)
        {
            // Search smallest element in the next row which
            // is greater than or equal to the current element
            int p = bsearch(0, m-1, arr[i][j], arr[i + 1]);
            ans = min(ans, abs(arr[i + 1][p] - arr[i][j]));

            // largest element which is smaller than the current
            // element in the next row must be just before
            // smallest element which is greater than or equal
            // to the current element because rows are sorted.
            if (p-1 >= 0)
                ans = min(ans, abs(arr[i + 1][p - 1] - arr[i][j]));
        }
    }
    return ans;
}

// Driven Program
int main()
{
    int m[R][C] =
    {
        8, 5,
        6, 8,
    };

    cout << mindiff(m, R, C) << endl;
    return 0;
}

Python

# Python program to find the minimum absolute difference
# between any of the adjacent elements of an array
# which is created by picking one element from each
# row of the matrix.
# R 2
# C 2
 
# Return smallest element greater than or equal
# to the current element.
def bsearch(low, high, n, arr):
    mid = (low + high)/2
 
    if(low <= high):
        if(arr[mid] < n):
            return bsearch(mid +1, high, n, arr);
        return bsearch(low, mid - 1, n, arr);
 
    return low;
 
# Return the minimum absolute difference adjacent
# elements of array
def mindiff(arr, n, m):

    # arr = [0 for i in range(R)][for j in range(C)]
    # Sort each row of the matrix.
    for i in range(n):
        sorted(arr)
 
    ans = 2147483647
 
    # For each matrix element
    for i in range(n-1):
        for j in range(m):
            # Search smallest element in the next row which
            # is greater than or equal to the current element
            p = bsearch(0, m-1, arr[i][j], arr[i + 1])
            ans = min(ans, abs(arr[i + 1][p] - arr[i][j]))
 
            # largest element which is smaller than the current
            # element in the next row must be just before
            # smallest element which is greater than or equal
            # to the current element because rows are sorted.
            if (p-1 >= 0):
                ans = min(ans, abs(arr[i + 1][p - 1] - arr[i][j]))
    return ans;
 
# Driver Program
m =[8, 5], [6, 8]
print mindiff(m, 2, 2)

# Contributed by: Afzal


Output:

0

Time Complexity : O(N*M*logM).

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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