# Divide an array into k segments to maximize maximum of segment minimums

Given an array of n integers, divide it into k segments and find the maximum of the minimums of k segments. Output the maximum integer that can be obtained among all ways to segment in k subarrays.

Examples:

```Input : arr[] = {1, 2, 3, 6, 5}
k = 2
Output: 5
Explanation: There are many ways to create
two segments. The optimal segments are (1, 2, 3)
and (6, 5). Minimum of both segments are 1 and 5,
hence the maximum(1, 5) is 5.

Input: -4 -5 -3 -2 -1 k=1
Output: -5
Explanation: only one segment, so minimum is -5.
```

There will be 3 cases that need to be considered.

1. k >= 3: When k is greater then 2, one segment will only compose of {max element}, so that max of minimum segments will always be the max.
2. k = 2: For k = 2 the answer is the maximum of the first and last element.
3. k = 1: Only possible partition is one segment equal to the whole array. So the answer is the minimum value on the whole array.

Below is the c++ implementation of the above approach

```// CPP Program to find maximum value of
// maximum of minimums of k segments.
#include <bits/stdc++.h>
using namespace std;

// function to calculate the max of all the
// minimum segments
int maxOfSegmentMins(int a[], int n, int k)
{
// if we have to divide it into 1 segment
// then the min will be the answer
if (k == 1)
return *min_element(a, a+n);

if (k == 2)
return max(a[0], a[n-1]);

// If k >= 3, return maximum of all
// elements.
return *max_element(a, a+n);
}

// driver program to test the above function
int main()
{
int a[] = { -10, -9, -8, 2, 7, -6, -5 };
int n = sizeof(a) / sizeof(a[0]);
int k = 2;
cout << maxOfSegmentMins(a, n, k);
}
```

Output:

`-5`

Time complexity: O(n)
Auxiliary Space: O(1)

Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits: http://www.geeksforgeeks.org
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