Maximise minimum element possible in Array after performing given operations
Last Updated :
23 Jul, 2025
Given an array arr[] of size N. The task is to maximize the minimum value of the array after performing given operations. In an operation, value x can be chosen and
- A value 3 * x can be subtracted from the arr[i] element.
- A value x is added to arr[i-1]. and
- A value of 2 * x can be added to arr[i-2].
Find the maximum possible minimum element of the array after any such operations.
Examples:
Input: arr[] = {1, 2, 3, 4, 5, 6}
Output: 3
Explanation: The last element is chosen and x =1 can be chosen.
So 3*x gets subtracted from arr[i] and x gets added to arr[i-1] and 2*x to arr[i-2] so the array becomes {1, 2, 3, 6, 6, 3}
In the 4th index x =1 can be chosen and now the array becomes {1, 2, 5, 7, 3, 3}.
In the 3rd index x = 1 can be chosen and now the array becomes {1, 4, 6, 4, 3, 3}.
In the 2nd index again x =1 can be chosen and now the array becomes {3, 4, 3, 4, 3, 3, 3}.
Hence the maximum possible minimum value is 3.
Input: arr[] = {9, 13, 167}
Output: 51
Naive Approach: This problem can be solved by checking for the possibility of maximum possible minimum value from [1, max(array)] by performing the operations from the end of the array.
Time Complexity: O(N2)
Space Complexity: O(1)
Efficient Approach: The efficient approach of this problem is based on Binary Search. Since it is based on maximizing the minimum value so by applying the binary search in the range [1, max(array)] and checking if mid is possible as a minimum element by performing the operations on the array such that every element is >=mid. Follow the steps below to solve the given problem:
- Initialize f = 1 and l = maximum element of the array and the res as INT_MIN.
- Perform binary search while f<=l
- Check if mid can be the minimum element by performing operations in the is_possible_min() function.
- In the is_possible_min() function
- Traverse from the end of the array (N-1) till index 2 and check if arr[i]<mid if it true return 0.
- Else find the extra which is 3x that can be added to arr[i-1] as x and arr[i-2] as 2x.
- If arr[0] >=mid and arr[1] >=mid return 1.
- Else return 0.
- If the is_possible_min() function returns true then mid is possible as the minimum value store the max(res, mid) in the res variable, so maximize the minimum value by moving right as f=mid +1
- Else move towards left and try if it is possible by l = mid -1.
- Print the res.
Below is the implementation of the above approach.
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Check if mid is possible as
// a minimum element after
// any number of operations using
// this predicate function
bool is_possible_min(vector<int> arr,
int mid)
{
int N = arr.size();
// Traverse from the end
for (int i = N - 1; i >= 2; i--) {
// mid can't be minimum
if (arr[i] < mid)
return 0;
else {
// Find the 3x
int extra = arr[i] - mid;
// Find the x
extra /= 3;
// Add x to a[i-1]
arr[i - 1] += extra;
// Add 2x to a[i-2]
arr[i - 2] += 2 * extra;
}
}
// Check if the first two elements
// are >= mid because if every element
// is greater than or equal to
// mid we can conclude
// mid as a minimum element
if (arr[0] >= mid && arr[1] >= mid)
return 1;
return 0;
}
// Function to find the
// maximum possible minimum value
void find_maximum_min(vector<int> arr)
{
// Initialize f = 1 and l as the
// maximum element of the array
int f = 1, l = *max_element(arr.begin(),
arr.end());
// Initialize res as INT_MIN
int res = INT_MIN;
// Perform binary search while f<=l
while (f <= l) {
int mid = (f + l) / 2;
// Check if mid is possible
// as a minimum element
if (is_possible_min(arr, mid)) {
// Take the max value of mid
res = max(res, mid);
// Try maximizing the min value
f = mid + 1;
}
// Move left if it is not possible
else {
l = mid - 1;
}
}
// Print the result
cout << res << endl;
}
// Driver Code
int main()
{
// Initialize the array
vector<int> arr = { 1, 2, 3, 4, 5, 6 };
// Function call
find_maximum_min(arr);
return 0;
}
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG {
// Check if mid is possible as
// a minimum element after
// any number of operations using
// this predicate function
static Boolean is_possible_min(int arr[],
int mid)
{
int N = arr.length;
// Traverse from the end
for (int i = N - 1; i >= 2; i--) {
// mid can't be minimum
if (arr[i] < mid)
return false;
else {
// Find the 3x
int extra = arr[i] - mid;
// Find the x
extra /= 3;
// Add x to a[i-1]
arr[i - 1] += extra;
// Add 2x to a[i-2]
arr[i - 2] += 2 * extra;
}
}
// Check if the first two elements
// are >= mid because if every element
// is greater than or equal to
// mid we can conclude
// mid as a minimum element
if (arr[0] >= mid && arr[1] >= mid)
return true;
return false;
}
// Function to find the
// maximum possible minimum value
static void find_maximum_min(int arr[])
{
// Initialize f = 1 and l as the
// maximum element of the array
int f = 1, l = Arrays.stream(arr).max().getAsInt();
// Initialize res as INT_MIN
int res = Integer.MIN_VALUE;
// Perform binary search while f<=l
while (f <= l) {
int mid = l + (f - l) / 2;
// Check if mid is possible
// as a minimum element
if (is_possible_min(arr, mid) == true) {
// Take the max value of mid
res = Math.max(res, mid);
// Try maximizing the min value
f = mid + 1;
}
// Move left if it is not possible
else {
l = mid - 1;
}
}
// Print the result
System.out.println(res);
}
// Driver Code
public static void main (String[] args)
{
// Initialize the array
int arr[] = { 1, 2, 3, 4, 5, 6 };
// Function call
find_maximum_min(arr);
}
}
// This code is contributed by hrithikgarg03188.
# python3 program for the above approach
INT_MIN = -2147483647 - 1
# Check if mid is possible as
# a minimum element after
# any number of operations using
# this predicate function
def is_possible_min(arr, mid):
N = len(arr)
# Traverse from the end
for i in range(N-1, 1, -1):
# mid can't be minimum
if (arr[i] < mid):
return 0
else:
# Find the 3x
extra = arr[i] - mid
# Find the x
extra //= 3
# Add x to a[i-1]
arr[i - 1] += extra
# Add 2x to a[i-2]
arr[i - 2] += 2 * extra
# Check if the first two elements
# are >= mid because if every element
# is greater than or equal to
# mid we can conclude
# mid as a minimum element
if (arr[0] >= mid and arr[1] >= mid):
return 1
return 0
# Function to find the
# maximum possible minimum value
def find_maximum_min(arr):
# Initialize f = 1 and l as the
# maximum element of the array
f, l = 1, max(arr)
# Initialize res as INT_MIN
res = INT_MIN
# Perform binary search while f<=l
while (f <= l):
mid = (f + l) // 2
# print(is_possible_min(arr,mid))
# Check if mid is possible
# as a minimum element
if (is_possible_min(arr.copy(), mid)):
# Take the max value of mid
res = max(res, mid)
# Try maximizing the min value
f = mid + 1
# Move left if it is not possible
else:
l = mid - 1
# Print the result
print(res)
# Driver Code
if __name__ == "__main__":
# Initialize the array
arr = [1, 2, 3, 4, 5, 6]
# Function call
find_maximum_min(arr)
# This code is contributed by rakeshsahni
// C# program for the above approach
using System;
using System.Linq;
class GFG
{
// Check if mid is possible as
// a minimum element after
// any number of operations using
// this predicate function
static bool is_possible_min(int[] arr, int mid)
{
int N = arr.Length;
// Traverse from the end
for (int i = N - 1; i >= 2; i--) {
// mid can't be minimum
if (arr[i] < mid)
return false;
else {
// Find the 3x
int extra = arr[i] - mid;
// Find the x
extra /= 3;
// Add x to a[i-1]
arr[i - 1] += extra;
// Add 2x to a[i-2]
arr[i - 2] += 2 * extra;
}
}
// Check if the first two elements
// are >= mid because if every element
// is greater than or equal to
// mid we can conclude
// mid as a minimum element
if (arr[0] >= mid && arr[1] >= mid)
return true;
return false;
}
// Function to find the
// maximum possible minimum value
static void find_maximum_min(int[] arr)
{
// Initialize f = 1 and l as the
// maximum element of the array
int f = 1, l = arr.Max();
// Initialize res as INT_MIN
int res = Int32.MinValue;
// Perform binary search while f<=l
while (f <= l) {
int mid = l + (f - l) / 2;
// Check if mid is possible
// as a minimum element
if (is_possible_min(arr, mid) == true) {
// Take the max value of mid
res = Math.Max(res, mid);
// Try maximizing the min value
f = mid + 1;
}
// Move left if it is not possible
else {
l = mid - 1;
}
}
// Print the result
Console.WriteLine(res);
}
// Driver Code
public static void Main()
{
// Initialize the array
int[] arr = { 1, 2, 3, 4, 5, 6 };
// Function call
find_maximum_min(arr);
}
}
// This code is contributed by Taranpreet
<script>
// Javascript program for the above approach
// Check if mid is possible as
// a minimum element after
// any number of operations using
// this predicate function
function is_possible_min(arr, mid) {
let N = arr.length;
// Traverse from the end
for (let i = N - 1; i >= 2; i--) {
// mid can't be minimum
if (arr[i] < mid)
return 0;
else {
// Find the 3x
let extra = arr[i] - mid;
// Find the x
extra = Math.floor(extra / 3);
// Add x to a[i-1]
arr[i - 1] += extra;
// Add 2x to a[i-2]
arr[i - 2] += 2 * extra;
}
}
// Check if the first two elements
// are >= mid because if every element
// is greater than or equal to
// mid we can conclude
// mid as a minimum element
if (arr[0] >= mid && arr[1] >= mid)
return 1;
return 0;
}
// Function to find the
// maximum possible minimum value
function find_maximum_min(arr) {
// Initialize f = 1 and l as the
// maximum element of the array
let f = 1, l = max_element(arr);
// Initialize res as INT_MIN
let res = Number.MIN_SAFE_INTEGER
// Perform binary search while f<=l
while (f <= l) {
let mid = Math.ceil((f + l) / 2);
// Check if mid is possible
// as a minimum element
if (is_possible_min(arr, mid)) {
// Take the max value of mid
res = Math.max(res, mid);
// Try maximizing the min value
f = mid + 1;
}
// Move left if it is not possible
else {
l = mid - 1;
}
}
// Print the result
document.write(res);
}
function max_element(ar) {
return [...ar].sort((a, b) => - a + b)[0]
}
// Driver Code
// Initialize the array
let arr = [1, 2, 3, 4, 5, 6];
// Function call
find_maximum_min(arr);
// This code is contributed by saurabh_jaiswal.
</script>
Time Complexity: O(N* log(maxval)) where maxval is the maximum element of the array. As we are using a while loop to traverse log(maxval) times as we are decrementing by floor division of 2 every traversal, and is_possible_min function will cost O (N) as we are using a loop to traverse N times.
Auxiliary Space: O(1), as we are not using any extra space.
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