# Find the maximum possible value of the minimum value of modified array

Given an array of size and a number . The task is to modify the given array such that:

• The difference between the sum of the array elements before and after modification is exactly equal to S.
• Modified array elements should be non-negative.
• The minimum value in the modified array must be maximized.
• To modify the given array, you can increment or decrement any element of the array.

The task is to find the minimum number of the modified array. If it is not possible then print -1. The minimum number should be as maximum as possible.

Examples:

```Input : a[] = {2, 2, 3}, S = 1
Output : 2
Explanation : Modified array is {2, 2, 2}

Input : a[] = {1, 3, 5}, S = 10
Output : -1
```

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

An efficient approach is to make a binary search between the minimum and the maximum possible value of the minimum number in a modified array. The minimum possible value is zero and the maximum possible array is minimum number in a given array. If given array elements sum is less than S then answer is not possible. so, print -1. If given array elements sum equals to S then answer will be zero.

Below is the implementation of the above approach:

## C++

 `// CPP program to find the maximum possible ` `// value of the minimum value of ` `// modified array ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find the maximum possible value ` `// of the minimum value of the modified array ` `int` `maxOfMin(``int` `a[], ``int` `n, ``int` `S) ` `{ ` `    ``// To store minimum value of array ` `    ``int` `mi = INT_MAX; ` ` `  `    ``// To store sum of elements of array ` `    ``int` `s1 = 0; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``s1 += a[i]; ` `        ``mi = min(a[i], mi); ` `    ``} ` ` `  `    ``// Solution is not possible ` `    ``if` `(s1 < S) ` `        ``return` `-1; ` ` `  `    ``// zero is the possible value ` `    ``if` `(s1 == S) ` `        ``return` `0; ` ` `  `    ``// minimum possible value ` `    ``int` `low = 0; ` ` `  `    ``// maximum possible value ` `    ``int` `high = mi; ` ` `  `    ``// to store a required answer ` `    ``int` `ans; ` ` `  `    ``// Binary Search ` `    ``while` `(low <= high) { ` ` `  `        ``int` `mid = (low + high) / 2; ` ` `  `        ``// If mid is possible then try to increase ` `        ``// required answer ` `        ``if` `(s1 - (mid * n) >= S) { ` `            ``ans = mid; ` `            ``low = mid + 1; ` `        ``} ` ` `  `        ``// If mid is not possible then decrease ` `        ``// required answer ` `        ``else` `            ``high = mid - 1; ` `    ``} ` ` `  `    ``// Return required answer ` `    ``return` `ans; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `a[] = { 10, 10, 10, 10, 10 }; ` ` `  `    ``int` `S = 10; ` ` `  `    ``int` `n = ``sizeof``(a) / ``sizeof``(a); ` ` `  `    ``cout << maxOfMin(a, n, S); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java  program to find the maximum possible ` `// value of the minimum value of ` `// modified array ` ` `  `import` `java.io.*; ` ` `  `class` `GFG { ` `     `  `// Function to find the maximum possible value ` `// of the minimum value of the modified array ` `static` `int` `maxOfMin(``int` `a[], ``int` `n, ``int` `S) ` `{ ` `    ``// To store minimum value of array ` `    ``int` `mi = Integer.MAX_VALUE; ` ` `  `    ``// To store sum of elements of array ` `    ``int` `s1 = ``0``; ` ` `  `    ``for` `(``int` `i = ``0``; i < n; i++) { ` `        ``s1 += a[i]; ` `        ``mi = Math.min(a[i], mi); ` `    ``} ` ` `  `    ``// Solution is not possible ` `    ``if` `(s1 < S) ` `        ``return` `-``1``; ` ` `  `    ``// zero is the possible value ` `    ``if` `(s1 == S) ` `        ``return` `0``; ` ` `  `    ``// minimum possible value ` `    ``int` `low = ``0``; ` ` `  `    ``// maximum possible value ` `    ``int` `high = mi; ` ` `  `    ``// to store a required answer ` `    ``int` `ans=``0``; ` ` `  `    ``// Binary Search ` `    ``while` `(low <= high) { ` ` `  `        ``int` `mid = (low + high) / ``2``; ` ` `  `        ``// If mid is possible then try to increase ` `        ``// required answer ` `        ``if` `(s1 - (mid * n) >= S) { ` `            ``ans = mid; ` `            ``low = mid + ``1``; ` `        ``} ` ` `  `        ``// If mid is not possible then decrease ` `        ``// required answer ` `        ``else` `            ``high = mid - ``1``; ` `    ``} ` ` `  `    ``// Return required answer ` `    ``return` `ans; ` `} ` ` `  `// Driver Code ` `    ``public` `static` `void` `main (String[] args) { ` ` `  `    ``int` `a[] = { ``10``, ``10``, ``10``, ``10``, ``10` `}; ` ` `  `    ``int` `S = ``10``; ` ` `  `    ``int` `n = a.length; ` ` `  `    ``System.out.println( maxOfMin(a, n, S)); ` `    ``} ` `//This code is contributed by ajit.     ` `} `

## Python

 `# Python program to find the maximum possible ` `# value of the minimum value of ` `# modified array ` ` `  ` `  `# Function to find the maximum possible value ` `# of the minimum value of the modified array ` `def` `maxOfMin(a, n, S): ` ` `  `    ``# To store minimum value of array ` `    ``mi ``=` `10``*``*``9` ` `  `    ``# To store sum of elements of array ` `    ``s1 ``=` `0` ` `  `    ``for` `i ``in` `range``(n): ` `        ``s1 ``+``=` `a[i] ` `        ``mi ``=` `min``(a[i], mi) ` `     `  ` `  `    ``# Solution is not possible ` `    ``if` `(s1 < S): ` `        ``return` `-``1` ` `  `    ``# zero is the possible value ` `    ``if` `(s1 ``=``=` `S): ` `        ``return` `0` ` `  `    ``# minimum possible value ` `    ``low ``=` `0` ` `  `    ``# maximum possible value ` `    ``high ``=` `mi ` ` `  `    ``# to store a required answer ` `    ``ans``=``0` ` `  `    ``# Binary Search ` `    ``while` `(low <``=` `high): ` ` `  `        ``mid ``=` `(low ``+` `high) ``/``/` `2` ` `  `        ``# If mid is possible then try to increase ` `        ``# required answer ` `        ``if` `(s1 ``-` `(mid ``*` `n) >``=` `S): ` `            ``ans ``=` `mid ` `            ``low ``=` `mid ``+` `1` `         `  ` `  `        ``# If mid is not possible then decrease ` `        ``# required answer ` `        ``else``: ` `            ``high ``=` `mid ``-` `1` `     `  ` `  `    ``# Return required answer ` `    ``return` `ans ` ` `  ` `  `# Driver Code ` ` `  `a``=``[``10``, ``10``, ``10``, ``10``, ``10``]  ` ` `  `S ``=` `10` ` `  `n ``=``len``(a) ` ` `  `print``(maxOfMin(a, n, S)) ` `#This code is contributed by Mohit kumar 29 `

## C#

 `// C# program to find the maximum possible  ` `// value of the minimum value of  ` `// modified array  ` ` `  `using` `System; ` ` `  `class` `GFG {  ` `     `  `    ``// Function to find the maximum possible value  ` `    ``// of the minimum value of the modified array  ` `    ``static` `int` `maxOfMin(``int` `[]a, ``int` `n, ``int` `S)  ` `    ``{  ` `        ``// To store minimum value of array  ` `        ``int` `mi = ``int``.MaxValue;  ` `     `  `        ``// To store sum of elements of array  ` `        ``int` `s1 = 0;  ` `     `  `        ``for` `(``int` `i = 0; i < n; i++) {  ` `            ``s1 += a[i];  ` `            ``mi = Math.Min(a[i], mi);  ` `        ``}  ` `     `  `        ``// Solution is not possible  ` `        ``if` `(s1 < S)  ` `            ``return` `-1;  ` `     `  `        ``// zero is the possible value  ` `        ``if` `(s1 == S)  ` `            ``return` `0;  ` `     `  `        ``// minimum possible value  ` `        ``int` `low = 0;  ` `     `  `        ``// maximum possible value  ` `        ``int` `high = mi;  ` `     `  `        ``// to store a required answer  ` `        ``int` `ans=0;  ` `     `  `        ``// Binary Search  ` `        ``while` `(low <= high) {  ` `     `  `            ``int` `mid = (low + high) / 2;  ` `     `  `            ``// If mid is possible then try to increase  ` `            ``// required answer  ` `            ``if` `(s1 - (mid * n) >= S) {  ` `                ``ans = mid;  ` `                ``low = mid + 1;  ` `            ``}  ` `     `  `            ``// If mid is not possible then decrease  ` `            ``// required answer  ` `            ``else` `                ``high = mid - 1;  ` `        ``}  ` `     `  `        ``// Return required answer  ` `        ``return` `ans;  ` `    ``}  ` ` `  `    ``// Driver Code  ` `    ``public` `static` `void` `Main () {  ` ` `  `    ``int` `[]a = { 10, 10, 10, 10, 10 };  ` ` `  `    ``int` `S = 10;  ` ` `  `    ``int` `n = a.Length;  ` ` `  `    ``Console.WriteLine(maxOfMin(a, n, S));  ` `    ``}  ` `    ``//This code is contributed by Ryuga  ` `}  `

## PHP

 `= ``\$S``)  ` `        ``{ ` `            ``\$ans` `= ``\$mid``; ` `            ``\$low` `= ``\$mid` `+ 1; ` `        ``} ` ` `  `        ``// If mid is not possible then ` `        ``// decrease required answer ` `        ``else` `            ``\$high` `= ``\$mid` `- 1; ` `    ``} ` ` `  `    ``// Return required answer ` `    ``return` `\$ans``; ` `} ` ` `  `// Driver Code ` `\$a` `= ``array``( 10, 10, 10, 10, 10 ); ` `\$S` `= 10; ` `\$n` `= sizeof(``\$a``); ` `echo` `maxOfMin(``\$a``, ``\$n``, ``\$S``); ` ` `  `// This code is contributed by akt_mit ` `?> `

Output:

```8
``` My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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