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

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[0]);       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 ?>

Javascript



Output:

8

Time Complexity: O(n + logn)

Auxiliary Space: O(1)

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