# Minimum value of maximum absolute difference of all adjacent pairs in an Array

Given an array arr, containing non-negative integers and (-1)s, of size N, the task is to replace those (-1)s with a common non-negative integer such that the maximum absolute difference of all adjacent pairs is minimum. Print this minimum possible value of the maximum absolute difference.

Examples:

Input: arr = {-1, -1, 11, -1, 3, -1}
Output: 4
Replace every -1 element with 7. Now the maximum absolute difference of all adjacent pairs is minimum which is equal to 4

Input: arr = {4, -1}
Output: 0

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

Approach:

1. Consider only those non-missing elements that are adjacent to at least one missing element.
2. Find the maximum element and the minimum element among them.
3. We need to find a value that minimizes the maximum absolute difference between the common value and these values.
4. The optimal value is equals to
`(minimum element + maximum element) / 2`

Below is the implementation of the above approach:

 `// C++ program to find the minimum value ` `// of maximum absolute difference of ` `// all adjacent pairs in an Array ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the minimum possible ` `// value of the maximum absolute difference. ` `int` `maximumAbsolute(``int` `arr[], ``int` `n) ` `{ ` `    ``// To store minimum and maximum elements ` `    ``int` `mn = INT_MAX; ` `    ``int` `mx = INT_MIN; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``// If right side element is equals -1 ` `        ``// and left side is not equals -1 ` `        ``if` `(i > 0 ` `            ``&& arr[i] == -1 ` `            ``&& arr[i - 1] != -1) { ` `            ``mn = min(mn, arr[i - 1]); ` `            ``mx = max(mx, arr[i - 1]); ` `        ``} ` ` `  `        ``// If left side element is equals -1 ` `        ``// and right side is not equals -1 ` `        ``if` `(i < n - 1 ` `            ``&& arr[i] == -1 ` `            ``&& arr[i + 1] != -1) { ` `            ``mn = min(mn, arr[i + 1]); ` `            ``mx = max(mx, arr[i + 1]); ` `        ``} ` `    ``} ` ` `  `    ``// Calculating the common integer ` `    ``// which needs to be replaced with ` `    ``int` `common_integer = (mn + mx) / 2; ` ` `  `    ``// Replace all -1 elements ` `    ``// with the common integer ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(arr[i] == -1) ` `            ``arr[i] = common_integer; ` `    ``} ` ` `  `    ``int` `max_diff = 0; ` ` `  `    ``// Calculating the maximum ` `    ``// absolute difference ` `    ``for` `(``int` `i = 0; i < n - 1; i++) { ` `        ``int` `diff = ``abs``(arr[i] - arr[i + 1]); ` ` `  `        ``if` `(diff > max_diff) ` `            ``max_diff = diff; ` `    ``} ` ` `  `    ``// Return the maximum absolute difference ` `    ``return` `max_diff; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { -1, -1, 11, -1, 3, -1 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` ` `  `    ``// Function call ` `    ``cout << maximumAbsolute(arr, n); ` ` `  `    ``return` `0; ` `} `

 `// Java program to find the minimum value ` `// of maximum absolute difference of ` `// all adjacent pairs in an Array ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `  `  `// Function to find the minimum possible ` `// value of the maximum absolute difference. ` `static` `int` `maximumAbsolute(``int` `arr[], ``int` `n) ` `{ ` `    ``// To store minimum and maximum elements ` `    ``int` `mn = Integer.MAX_VALUE; ` `    ``int` `mx = Integer.MIN_VALUE; ` `  `  `    ``for` `(``int` `i = ``0``; i < n; i++) { ` ` `  `        ``// If right side element is equals -1 ` `        ``// and left side is not equals -1 ` `        ``if` `(i > ``0` `            ``&& arr[i] == -``1` `            ``&& arr[i - ``1``] != -``1``) { ` `            ``mn = Math.min(mn, arr[i - ``1``]); ` `            ``mx = Math.max(mx, arr[i - ``1``]); ` `        ``} ` `  `  `        ``// If left side element is equals -1 ` `        ``// and right side is not equals -1 ` `        ``if` `(i < n - ``1` `            ``&& arr[i] == -``1` `            ``&& arr[i + ``1``] != -``1``) { ` `            ``mn = Math.min(mn, arr[i + ``1``]); ` `            ``mx = Math.max(mx, arr[i + ``1``]); ` `        ``} ` `    ``} ` `  `  `    ``// Calculating the common integer ` `    ``// which needs to be replaced with ` `    ``int` `common_integer = (mn + mx) / ``2``; ` `  `  `    ``// Replace all -1 elements ` `    ``// with the common integer ` `    ``for` `(``int` `i = ``0``; i < n; i++) { ` `        ``if` `(arr[i] == -``1``) ` `            ``arr[i] = common_integer; ` `    ``} ` `  `  `    ``int` `max_diff = ``0``; ` `  `  `    ``// Calculating the maximum ` `    ``// absolute difference ` `    ``for` `(``int` `i = ``0``; i < n - ``1``; i++) { ` `        ``int` `diff = Math.abs(arr[i] - arr[i + ``1``]); ` `  `  `        ``if` `(diff > max_diff) ` `            ``max_diff = diff; ` `    ``} ` `  `  `    ``// Return the maximum absolute difference ` `    ``return` `max_diff; ` `} ` `  `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { -``1``, -``1``, ``11``, -``1``, ``3``, -``1` `}; ` `    ``int` `n = arr.length; ` `  `  `    ``// Function call ` `    ``System.out.print(maximumAbsolute(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

 `# Python3 program to find the minimum value ` `# of maximum absolute difference of ` `# all adjacent pairs in an Array ` ` `  `# Function to find the minimum possible ` `# value of the maximum absolute difference. ` `def` `maximumAbsolute(arr, n): ` ` `  `    ``# To store minimum and maximum elements ` `    ``mn ``=` `10``*``*``9` `    ``mx ``=` `-``10``*``*``9` ` `  `    ``for` `i ``in` `range``(n): ` `     `  `        ``# If right side element is equals -1 ` `        ``# and left side is not equals -1 ` `        ``if` `(i > ``0` `            ``and` `arr[i] ``=``=` `-``1` `            ``and` `arr[i ``-` `1``] !``=` `-``1``): ` `            ``mn ``=` `min``(mn, arr[i ``-` `1``]) ` `            ``mx ``=` `max``(mx, arr[i ``-` `1``]) ` ` `  `        ``# If left side element is equals -1 ` `        ``# and right side is not equals -1 ` `        ``if` `(i < n ``-` `1` `            ``and` `arr[i] ``=``=` `-``1` `            ``and` `arr[i ``+` `1``] !``=` `-``1``): ` `            ``mn ``=` `min``(mn, arr[i ``+` `1``]) ` `            ``mx ``=` `max``(mx, arr[i ``+` `1``]) ` ` `  `    ``# Calculating the common integer ` `    ``# which needs to be replaced with ` `    ``common_integer ``=` `(mn ``+` `mx) ``/``/` `2` ` `  `    ``# Replace all -1 elements ` `    ``# with the common integer ` `    ``for` `i ``in` `range``(n): ` `        ``if` `(arr[i] ``=``=` `-``1``): ` `            ``arr[i] ``=` `common_integer ` ` `  `    ``max_diff ``=` `0` ` `  `    ``# Calculating the maximum ` `    ``# absolute difference ` `    ``for` `i ``in` `range``(n``-``1``): ` `        ``diff ``=` `abs``(arr[i] ``-` `arr[i ``+` `1``]) ` ` `  `        ``if` `(diff > max_diff): ` `            ``max_diff ``=` `diff ` ` `  `    ``# Return the maximum absolute difference ` `    ``return` `max_diff ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr``=``[``-``1``, ``-``1``, ``11``, ``-``1``, ``3``, ``-``1``] ` `    ``n ``=` `len``(arr) ` ` `  `    ``# Function call ` `    ``print``(maximumAbsolute(arr, n)) ` ` `  `# This code is contributed by mohit kumar 29 `

 `// C# program to find the minimum value ` `// of maximum absolute difference of ` `// all adjacent pairs in an Array ` `using` `System; ` ` `  `class` `GFG{ ` `  `  `    ``// Function to find the minimum possible ` `    ``// value of the maximum absolute difference. ` `    ``static` `int` `maximumAbsolute(``int` `[]arr, ``int` `n) ` `    ``{ ` `        ``// To store minimum and maximum elements ` `        ``int` `mn = ``int``.MaxValue; ` `        ``int` `mx = ``int``.MinValue; ` `      `  `        ``for` `(``int` `i = 0; i < n; i++) { ` `     `  `            ``// If right side element is equals -1 ` `            ``// and left side is not equals -1 ` `            ``if` `(i > 0 ` `                ``&& arr[i] == -1 ` `                ``&& arr[i - 1] != -1) { ` `                ``mn = Math.Min(mn, arr[i - 1]); ` `                ``mx = Math.Max(mx, arr[i - 1]); ` `            ``} ` `      `  `            ``// If left side element is equals -1 ` `            ``// and right side is not equals -1 ` `            ``if` `(i < n - 1 ` `                ``&& arr[i] == -1 ` `                ``&& arr[i + 1] != -1) { ` `                ``mn = Math.Min(mn, arr[i + 1]); ` `                ``mx = Math.Max(mx, arr[i + 1]); ` `            ``} ` `        ``} ` `      `  `        ``// Calculating the common integer ` `        ``// which needs to be replaced with ` `        ``int` `common_integer = (mn + mx) / 2; ` `      `  `        ``// Replace all -1 elements ` `        ``// with the common integer ` `        ``for` `(``int` `i = 0; i < n; i++) { ` `            ``if` `(arr[i] == -1) ` `                ``arr[i] = common_integer; ` `        ``} ` `      `  `        ``int` `max_diff = 0; ` `      `  `        ``// Calculating the maximum ` `        ``// absolute difference ` `        ``for` `(``int` `i = 0; i < n - 1; i++) { ` `            ``int` `diff = Math.Abs(arr[i] - arr[i + 1]); ` `      `  `            ``if` `(diff > max_diff) ` `                ``max_diff = diff; ` `        ``} ` `      `  `        ``// Return the maximum absolute difference ` `        ``return` `max_diff; ` `    ``} ` `      `  `    ``// Driver Code ` `    ``public` `static` `void` `Main(``string``[] args) ` `    ``{ ` `        ``int` `[]arr = { -1, -1, 11, -1, 3, -1 }; ` `        ``int` `n = arr.Length; ` `      `  `        ``// Function call ` `        ``Console.Write(maximumAbsolute(arr, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by Yash_R `

Output:
```4
```

Time complexity: O(N)

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Improved By : mohit kumar 29, Rajput-Ji, Yash_R

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