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

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, element picked from row 2 will become arr and so on.

Examples:

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

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

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 ` `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; ` `} `

## Java

 `// Java 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 ` `import` `java.util.Arrays; ` `class` `GFG  ` `{ ` `static` `final` `int` `R=``2``; ` `static` `final` `int` `C=``2``; ` ` `  `// Return smallest element greater than  ` `// or equal to the current element. ` `static` `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 ` `static` `int` `mindiff(``int` `arr[][], ``int` `n, ``int` `m) ` `{ ` ` `  `    ``// Sort each row of the matrix. ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``Arrays.sort(arr[i]); ` ` `  `    ``int` `ans = +``2147483647``; ` ` `  `    ``// 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 = Math.min(ans, Math.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 = Math.min(ans,  ` `                        ``Math.abs(arr[i + ``1``][p - ``1``] - arr[i][j])); ` `        ``} ` `    ``} ` `    ``return` `ans; ` `} ` ` `  `    ``//Driver code ` `public` `static` `void` `main (String[] args) ` `{ ` `    ``int` `m[][] ={{``8``, ``5``}, ` `                ``{``6``, ``8``}}; ` ` `  `    ``System.out.println(mindiff(m, R, C)); ` `} ` `} ` `//This code is contributed by Anant Agarwal. `

## 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``) ` ` `  `# This code is contributed by Afzal `

## 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  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` `     `  `static` `public` `int` `R=2;  ` `static` `public` `int` `C=2;  ` ` `  `// Return smallest element greater than  ` `// or equal to the current element.  ` `static` `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;  ` `}  ` `public` `static` `int``[] GetRow(``int``[,] matrix, ``int` `row) ` `{ ` `    ``var` `rowLength = matrix.GetLength(1); ` `    ``var` `rowVector = ``new` `int``[rowLength]; ` ` `  `    ``for` `(``var` `i = 0; i < rowLength; i++) ` `    ``rowVector[i] = matrix[row, i]; ` ` `  `    ``return` `rowVector; ` `} ` `// Return the minimum absolute difference adjacent  ` `// elements of array  ` `static` `int` `mindiff(``int` `[,]arr, ``int` `n, ``int` `m)  ` `{  ` ` `  `    ``// Sort each row of the matrix.  ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `        ``Array.Sort(GetRow(arr,i));  ` ` `  `    ``int` `ans = +2147483647;  ` ` `  `    ``// 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], GetRow(arr,i+1));  ` `        ``ans = Math.Min(ans, Math.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 = Math.Min(ans,  ` `                        ``Math.Abs(arr[i + 1,p - 1] - arr[i,j]));  ` `        ``}  ` `    ``}  ` `    ``return` `ans;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `Main (String[] args)  ` `{  ` `    ``int` `[,]m ={{8, 5},  ` `                ``{6, 8}};  ` ` `  `    ``Console.WriteLine(mindiff(m, R, C));  ` `}  ` `}  ` ` `  `// This code is contributed by 29AjayKumar `

## PHP

 `= 0) ` `                ``\$ans` `= min(``\$ans``, ``abs``(``\$arr``[``\$i` `+ 1][``\$p` `- 1] -  ` `                                     ``\$arr``[``\$i``][``\$j``])); ` `        ``} ` `    ``} ` `    ``return` `\$ans``; ` `} ` ` `  `// Driver Code ` `\$m` `= ``array``(8, 5, 6, 8); ` ` `  `echo` `mindiff(``\$m``, ``\$R``, ``\$C``), ``"\n"``; ` ` `  `// This code is contributed by Sach_Code ` `?> `

Output:

```0
```

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

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