Maximize minimum element of an Array using operations

Last Updated : 31 Aug, 2023

Given an array A[] of N integers and two integers X and Y (X ≤ Y), the task is to find the maximum possible value of the minimum element in an array A[] of N integers by adding X to one element and subtracting Y from another element any number of times, where X ≤ Y.

Examples:

Input: N= 3, A[] = {1, 5, 9}, X = 2, Y = 2
Output: 5
Explanation: We will perform the operation twice as follows:

Add X to first element and subtract Y from third element. Array becomes - {3, 5, 7}
Again, add X to first element and subtract Y from third element. Array becomes - {5, 5, 5}
The minimum element is 5. It can be checked that after performing operation with any other indices or any more number of times, the minimum element cannot be made greater than 5.

Input: N = 3, A[] = {11, 1, 2}, X = 2, Y = 3
Output: 3
Explanation: We will perform the operation twice as follows:

Add X to second element and subtract Y from first element. Array becomes - {8, 3, 2}
Add X to third element and subtract Y from first element. Array becomes - {5, 3, 4}
The minimum element is 3. It can be checked that after performing operation with any other indices or any more number of times, the minimum element cannot be made greater than 3.

Approach: To solve the problem follow the below idea:

The idea is to use binary search. Initialize two variables lo and hi to the minimum and maximum element of the array, respectively. Now, calculate mid value between lo and hi using binary search. Iterate the array to count the number of additions and subtractions required to make all elements either greater than or equal to mid, using X and Y respectively. If the total number of subtractions required is greater than or equal to the total number of additions required, the update lo=mid and continues the binary search, else update hi=mid-1 and continue the search. When lo and hi become equal, return this value as the answer.

Below are the steps for the above approach:

Initialize the lower bound (say "lo") and upper bound(say "hi") for binary search with minimum and maximum elements of the array respectively.
In each iteration of the binary search, initialize two variables, say "add" and "subtract" with 0.
Iterate through the array.
If the current element of the array is greater than or equal to mid, add (A[i]-mid)/Y to "subtract".
Else, add (mid-A[i]+X-1)/X to "add".
If "subtract" is greater than or equal to "add", set lo=mid.
Else, set hi=mid-1.
When lo becomes greater than or equal to hi, return lo.

Below is the code for the above approach:

C++

// C++ code for the above approach #include <bits/stdc++.h> using namespace std; #define int long long  // Function to Maximize the minimum element // of array by adding and subtracting X and // Y from distinct elements any number // of times int maximizeMinimumElement(int N, int A[], int X, int Y) {      // Initializing lo and hi     int lo = *min_element(A, A + N);     int hi = *max_element(A, A + N);      while (lo < hi) {          // Initializing mid, add and         // subtract         int mid = lo + (hi - lo + 1) / 2;         int add = 0, subtract = 0;          // Iterating through the array         for (int i = 0; i < N; i++) {              // If mid is less than or equal             // to A[i], accordingly increase             // the value of subtract             if (mid <= A[i]) {                 subtract += (A[i] - mid) / Y;             }              // Else, accordingly increase             // the value of add             else {                 add += (mid - A[i] + X - 1) / X;             }         }          // If subtract is greater than or         // equal to add, set lo = mid         if (subtract >= add) {             lo = mid;         }          // Else set hi = mid-1         else {             hi = mid - 1;         }     }      // Return the answer     return lo; }  // Driver code int32_t main() {     int N = 3, X = 2, Y = 2;     int A[] = { 1, 5, 9 };      // Function Call     cout << maximizeMinimumElement(N, A, X, Y);     return 0; }

Java

// Java code for the above approach import java.util.*;  class GFG {      // Function to Maximize the minimum element     // of array by adding and subtracting X and     // Y from distinct elements any number     // of times     public static int maximizeMinimumElement(int N, int A[],                                              int X, int Y)     {          // Initializing lo and hi         int lo = Arrays.stream(A).min().getAsInt();         int hi = Arrays.stream(A).max().getAsInt();          while (lo < hi) {              // Initializing mid, add and             // subtract             int mid = lo + (hi - lo + 1) / 2;             int add = 0, subtract = 0;              // Iterating through the array             for (int i = 0; i < N; i++) {                  // If mid is less than or equal                 // to A[i], accordingly increase                 // the value of subtract                 if (mid <= A[i]) {                     subtract += (A[i] - mid) / Y;                 }                  // Else, accordingly increase                 // the value of add                 else {                     add += (mid - A[i] + X - 1) / X;                 }             }              // If subtract is greater than or             // equal to add, set lo = mid             if (subtract >= add) {                 lo = mid;             }              // Else set hi = mid-1             else {                 hi = mid - 1;             }         }          // Return the answer         return lo;     }      // Driver code     public static void main(String[] args)     {         int N = 3, X = 2, Y = 2;         int A[] = { 1, 5, 9 };          // Function Call         System.out.println(             maximizeMinimumElement(N, A, X, Y));     } } // This code is contributed by prasad264

Python3

def maximizeMinimumElement(N, A, X, Y):     # Initializing lo and hi     lo = min(A)     hi = max(A)      while lo < hi:         # Initializing mid, add and subtract         mid = lo + (hi - lo + 1) // 2         add = 0         subtract = 0          # Iterating through the array         for i in range(N):             # If mid is less than or equal to A[i],              # accordingly increase the value of subtract             if mid <= A[i]:                 subtract += (A[i] - mid) // Y             # Else, accordingly increase the value of add             else:                 add += (mid - A[i] + X - 1) // X          # If subtract is greater than or equal to add, set lo = mid         if subtract >= add:             lo = mid         # Else set hi = mid-1         else:             hi = mid - 1      # Return the answer     return lo  # Driver code N = 3 X = 2 Y = 2 A = [1, 5, 9]  # Function Call print(maximizeMinimumElement(N, A, X, Y))

using System;  class GFG {     static int MaximizeMinimumElement(int N, int[] A, int X, int Y)     {         // Initializing lo and hi         int lo = A[0];         int hi = A[0];                  // Find the minimum and maximum values in the array         for (int i = 1; i < N; i++)         {             lo = Math.Min(lo, A[i]);             hi = Math.Max(hi, A[i]);         }          while (lo < hi)         {             // Initializing mid, add and subtract             int mid = lo + (hi - lo + 1) / 2;             int add = 0;             int subtract = 0;              // Iterating through the array             for (int i = 0; i < N; i++)             {                 // If mid is less than or equal to A[i],                 // accordingly increase the value of subtract                 if (mid <= A[i])                 {                     subtract += (A[i] - mid) / Y;                 }                 // Else, accordingly increase the value of add                 else                 {                     add += (mid - A[i] + X - 1) / X;                 }             }              // If subtract is greater than or equal to add, set lo = mid             if (subtract >= add)             {                 lo = mid;             }             // Else set hi = mid-1             else             {                 hi = mid - 1;             }         }          // Return the answer         return lo;     }      static void Main(string[] args)     {         int N = 3;         int X = 2;         int Y = 2;         int[] A = new int[] { 1, 5, 9 };          // Function Call         Console.WriteLine(MaximizeMinimumElement(N, A, X, Y));     } }

JavaScript

// JavaScript code for the above approach  // Function to maximize the minimum element  // of an array by adding and subtracting X and  // Y from distinct elements any number of times function maximizeMinimumElement(N, A, X, Y) {          // Initializing lo and hi     let lo = Math.min(...A);     let hi = Math.max(...A);      while (lo < hi) {         // Initializing mid, add and          // subtract         let mid = lo + Math.floor((hi - lo + 1) / 2);         let add = 0;         let subtract = 0;          // Iterating through the array         for (let i = 0; i < N; i++) {             // If mid is less than or equal              // to A[i], accordingly increase              // the value of subtract             if (mid <= A[i]) {                 subtract += Math.floor((A[i] - mid) / Y);             }                          // Else, accordingly increase              // the value of add             else {                 add += Math.floor((mid - A[i] + X - 1) / X);             }         }          // If subtract is greater than or          // equal to add, set lo = mid         if (subtract >= add) {             lo = mid;         }                  // Else set hi = mid-1         else {             hi = mid - 1;         }     }      // Return the answer     return lo; }  // Driver code let N = 3; let X = 2; let Y = 2; let A = [1, 5, 9];  // Function call console.log(maximizeMinimumElement(N, A, X, Y));

Output

Time Complexity: O(N*logN)
Auxiliary Space: O(1)

Maximize the minimum element and return it

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Maximize minimum element of an Array using operations

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