Given an array a of size N and an integer K, the task is to divide the array into K segments such that sum of the minimum of K segments is maximized.
Examples:
Input: a[] = {5, 7, 4, 2, 8, 1, 6}, K = 3
Output: 7
Divide the array at indexes 0 and 1. Then the segments are {5}, {7}, {4, 2, 8, 1, 6}. Sum of the minimum is 5 + 7 + 1 = 13Input: a[] = {6, 5, 3, 8, 9, 10, 4, 7, 10}, K = 4
Output: 27
Approach: The problem can be solved using Dynamic Programming. Try all the possible partitions that are possible using recursion. Let dp[i][k] be the maximum sum of minimums till index i with k partitions. Hence the possible states will be partition at every index from the index i till n. The maximum sum of minimums of all those states will be our answer. After writing this recurrence, we can use memoization.
Below is the implementation of the above approach:
C++
// C++ program to find the sum of// the minimum of all the segments#include <bits/stdc++.h>using namespace std;const int MAX = 10;// Function to maximize the sum of the minimumsint maximizeSum(int a[], int n, int ind, int k, int dp[MAX][MAX]){ // If k segments have been divided if (k == 0) { // If we are at the end if (ind == n) return 0; // If we donot reach the end // then return a negative number // that cannot be the sum else return -1e9; } // If at the end but // k segments are not formed else if (ind == n) return -1e9; // If the state has not been visited yet else if (dp[ind][k] != -1) return dp[ind][k]; // If the state has not been visited else { int ans = 0; // Get the minimum element in the segment int mini = a[ind]; // Iterate and try to break at every index // and create a segment for (int i = ind; i < n; i++) { // Find the minimum element in the segment mini = min(mini, a[i]); // Find the sum of all the segments trying all // the possible combinations ans = max(ans, maximizeSum( a, n, i + 1, k - 1, dp) + mini); } // Return the answer by // memoizing it return dp[ind][k] = ans; }}// Driver Codeint main(){ int a[] = { 5, 7, 4, 2, 8, 1, 6 }; int k = 3; int n = sizeof(a) / sizeof(a[0]); // Initialize dp array with -1 int dp[MAX][MAX]; memset(dp, -1, sizeof dp); cout << maximizeSum(a, n, 0, k, dp); return 0;} |
Java
// Java program to find the sum of // the minimum of all the segmentsclass GFG { static int MAX = 10; // Function to maximize the // sum of the minimums public static int maximizeSum( int[] a, int n, int ind, int k, int[][] dp) { // If k segments have been divided if (k == 0) { // If we are at the end if (ind == n) return 0; // If we donot reach the end // then return a negative number // that cannot be the sum else return -1000000000; } // If at the end but // k segments are not formed else if (ind == n) return -1000000000; // If the state has not // been visited yet else if (dp[ind][k] != -1) return dp[ind][k]; // If the state has // not been visited else { int ans = 0; // Get the minimum // element in the segment int mini = a[ind]; // Iterate and try to break // at every index // and create a segment for (int i = ind; i < n; i++) { // Find the minimum element // in the segment mini = Math.min(mini, a[i]); // Find the sum of all the // segments trying all // the possible combinations ans = Math.max(ans, maximizeSum(a, n, i + 1, k - 1, dp) + mini); } // Return the answer by // memoizing it return dp[ind][k] = ans; } } // Driver Code public static void main(String[] args) { int[] a = { 5, 7, 4, 2, 8, 1, 6 }; int k = 3; int n = a.length; // Initialize dp array with -1 int[][] dp = new int[MAX][MAX]; for (int i = 0; i < MAX; i++) { for (int j = 0; j < MAX; j++) dp[i][j] = -1; } System.out.println( maximizeSum(a, n, 0, k, dp)); }}// This code is contributed by// sanjeev2552 |
Python3
# Python 3 program to find the sum of# the minimum of all the segmentsMAX = 10# Function to maximize the# sum of the minimumsdef maximizeSum(a,n, ind, k, dp): # If k segments have been divided if (k == 0): # If we are at the end if (ind == n): return 0 # If we donot reach the end # then return a negative number # that cannot be the sum else: return -1e9 # If at the end but # k segments are not formed elif (ind == n): return -1e9 # If the state has been visited already elif (dp[ind][k] != -1): return dp[ind][k] # If the state has not been visited else: ans = 0 # Get the minimum element # in the segment mini = a[ind] # Iterate and try to break # at every index # and create a segment for i in range(ind,n,1): # Find the minimum element # in the segment mini = min(mini, a[i]) # Find the sum of all the # segments trying all # the possible combinations ans = max(ans, maximizeSum(\ a, n, i + 1, k - 1, dp) + mini) # Return the answer by # memoizing it dp[ind][k] = ans return ans # Driver Codeif __name__ == '__main__': a = [5, 7, 4, 2, 1, 6] k = 3 n = len(a) # Initialize dp array with -1 dp = [[-1 for i in range(MAX)]\ for j in range(MAX)] print(maximizeSum(a, n, 0, k, dp))# This code is contributed by# Surendra_Gangwar |
C#
// C# program to find the sum of // the minimum of all the segmentsusing System; class GFG { static int MAX = 10; // Function to maximize the sum of the minimums public static int maximizeSum( int[] a, int n, int ind, int k, int[] dp) { // If k segments have been divided if (k == 0) { // If we are at the end if (ind == n) return 0; // If we donot reach the end // then return a negative number // that cannot be the sum else return -1000000000; } // If at the end but // k segments are not formed else if (ind == n) return -1000000000; // If the state has not // been visited yet else if (dp[ind, k] != -1) return dp[ind, k]; // If the state has not been visited else { int ans = 0; // Get the minimum element // in the segment int mini = a[ind]; // Iterate and try to break // at every index // and create a segment for (int i = ind; i < n; i++) { // Find the minimum element // in the segment mini = Math.Min(mini, a[i]); // Find the sum of all the // segments trying all // the possible combinations ans = Math.Max(ans, maximizeSum(a, n, i + 1, k - 1, dp) + mini); } // Return the answer by // memoizing it return dp[ind,k] = ans; } } // Driver Code public static void Main(String[] args) { int[] a = { 5, 7, 4, 2, 8, 1, 6 }; int k = 3; int n = a.Length; // Initialize dp array with -1 int[,] dp = new int[MAX, MAX]; for (int i = 0; i < MAX; i++) { for (int j = 0; j < MAX; j++) dp[i, j] = -1; } Console.WriteLine( maximizeSum(a, n, 0, k, dp)); }}// This code is contributed by 29AjayKumar |
Javascript
<script>// JavaScript program to find the sum of // the minimum of all the segments var MAX = 10; // Function to maximize the // sum of the minimums function maximizeSum(a , n , ind , k, dp) { // If k segments have been divided if (k == 0) { // If we are at the end if (ind == n) return 0; // If we donot reach the end // then return a negative number // that cannot be the sum else return -1000000000; } // If at the end but // k segments are not formed else if (ind == n) return -1000000000; // If the state has not // been visited yet else if (dp[ind][k] != -1) return dp[ind][k]; // If the state has // not been visited else { var ans = 0; // Get the minimum // element in the segment var mini = a[ind]; // Iterate and try to break // at every index // and create a segment for (i = ind; i < n; i++) { // Find the minimum element // in the segment mini = Math.min(mini, a[i]); // Find the sum of all the // segments trying all // the possible combinations ans = Math.max(ans, maximizeSum(a, n, i + 1, k - 1, dp) + mini); } // Return the answer by // memoizing it return dp[ind][k] = ans; } } // Driver Code var a = [ 5, 7, 4, 2, 8, 1, 6 ]; var k = 3; var n = a.length; // Initialize dp array with -1 var dp = Array(MAX).fill().map(()=>Array(MAX).fill(0)); for (var i = 0; i < MAX; i++) { for (j = 0; j < MAX; j++) dp[i][j] = -1; } document.write(maximizeSum(a, n, 0, k, dp));// This code contributed by Rajput-Ji </script> |
Output
13
Time Complexity: O(N * N * K)
Auxiliary Space: O(N * K)
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