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

Last Updated : 18 Sep, 2023

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 than 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 implementation of the above approach

C++

// 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); }

Java

// Java Program to find maximum  // value of maximum of minimums // of k segments. import java .io.*; import java .util.*;  class GFG {      // function to calculate      // the max of all the      // minimum segments     static 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)          {             Arrays.sort(a);                 return a[0];         }              if (k == 2)              return Math.max(a[0],                              a[n - 1]);                   // If k >= 3, return          // maximum of all          // elements.         return a[n - 1];     }          // Driver Code     static public void main (String[] args)     {         int []a = {-10, -9, -8,                      2, 7, -6, -5};         int n = a.length;         int k = 2;                  System.out.println(                 maxOfSegmentMins(a, n, k));     } }  // This code is contributed // by anuj_67.

Python3

# Python3 Program to find maximum value of  # maximum of minimums of k segments.  # function to calculate the max of all the  # minimum segments  def maxOfSegmentMins(a,n,k):      # if we have to divide it into 1 segment      # then the min will be the answer      if k ==1:         return min(a)     if k==2:         return max(a[0],a[n-1])      # If k >= 3, return maximum of all      #  elements.      return max(a)      # Driver code if __name__=='__main__':     a = [-10, -9, -8, 2, 7, -6, -5]     n = len(a)     k =2     print(maxOfSegmentMins(a,n,k))  # This code is contributed by  # Shrikant13

// C# Program to find maximum value of // maximum of minimums of k segments. using System; using System.Linq;  public class GFG {       // function to calculate the max     // of all the minimum segments     static 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 a.Min();              if (k == 2)              return Math.Max(a[0], a[n - 1]);                   // If k >= 3, return maximum of         // all elements.         return a.Max();     }          // Driver function     static public void Main ()     {         int []a = { -10, -9, -8, 2, 7,                                  -6, -5 };         int n = a.Length;         int k = 2;                  Console.WriteLine(                   maxOfSegmentMins(a, n, k));     } }  // This code is contributed by vt_m.

PHP

<?php // PHP Program to find maximum // value of maximum of minimums // of k segments.   // function to calculate // the max of all the // minimum segments function maxOfSegmentMins($a, $n, $k) {     // if we have to divide      // it into 1 segment then     // the min will be the answer     if ($k == 1)      return min($a);      if ($k == 2)      return max($a[0],                 $a[$n - 1]);           // If k >= 3, return      // maximum of all elements.     return max($a); }  // Driver Code $a = array(-10, -9, -8,              2, 7, -6, -5); $n = count($a); $k = 2; echo maxOfSegmentMins($a, $n, $k);  // This code is contributed by vits. ?>

JavaScript

<script> // javascript Program to find maximum  // value of maximum of minimums // of k segments.     // function to calculate     // the max of all the     // minimum segments     function maxOfSegmentMins(a , n , k)     {          // if we have to divide         // it into 1 segment then         // the min will be the answer         if (k == 1) {             a.sort();             return a[0];         }          if (k == 2)             return Math.max(a[0], a[n - 1]);          // If k >= 3, return         // maximum of all         // elements.         return a[n - 1];     }      // Driver Code         var a = [ -10, -9, -8, 2, 7, -6, -5 ];         var n = a.length;         var k = 2;          document.write(maxOfSegmentMins(a, n, k));  // This code is contributed by Rajput-Ji </script>

Output

-5

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

Divide the array in K segments such that the sum of minimums is maximized

Raja Vikramaditya

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

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