# 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.

**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.**k = 2:**For k = 2 the answer is the maximum of the first and last element.**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)

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source and credits.

source and credits: http://www.geeksforgeeks.org

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