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Write a program to Find the maximised sum of removed elements.You are given a stack of N integers such that the first element represents the top of the stack and the last element represents the bottom of the stack. You need to pop at least one element from the stack. At any one moment, you can convert stack into a queue. The bottom of the stack represents the front of the queue. You cannot convert the queue back into a stack. Your task is to remove exactly K elements such that the sum of the K removed elements is maximised.

Question

Write a program to Find the maximised sum of removed elements.You are given a stack of N integers such that the first element represents the top of the stack and the last element represents the bottom of the stack. You need to pop at least one element from the stack. At any one moment, you can convert stack into a queue. The bottom of the stack represents the front of the queue. You cannot convert the queue back into a stack. Your task is to remove exactly K elements such that the sum of the K removed elements is maximised.

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Solution

Here is a Python solution for the problem:

from collections import deque

def maxSum(stack, k):
    n = len(stack)
    dp = [[0 for _ in range(k+1)] for _ in range(n+1)]
    prefix_sum = [0 for _ in range(n+1)]
    queue = deque()

    for i in range(n-1, -1, -1):
        queue.appendleft(stack[i])
        prefix_sum[i] = prefix_sum[i+1] + stack[i]

    for i in range(1, n+1):
        for j in range(1, k+1):
            dp[i][j] = max(dp[i-1][j], dp[i-1][j-1] + queue[i-1])

    return dp[n][k]

stack = [1, 2, 3, 4, 5]
k = 2
print(maxSum(stack, k))  # Output: 9

This program uses dynamic programming to solve the problem. The dp array is used to store the maximum sum that can be obtained by removing j elements from the first i elements of the stack. The prefix_sum array is used to store the sum of all elements from the ith element to the end of the stack. The queue is used to store the elements of the stack in reverse order. The program iterates over all elements of the stack and for each element, it calculates the maximum sum that can be obtained by removing j elements from the first i elements of the stack. The maximum sum is either the maximum sum obtained by not removing the current element or the maximum sum obtained by removing the current element and adding its value to the maximum sum obtained by removing j-1 elements from the first i-1 elements of the stack. The maximum sum that can be obtained by removing k elements from the stack is stored in dp[n][k].

This problem has been solved

Similar Questions

Given an array of integers called nums, you can perform the following operation while nums contains at least 2 elements:Choose the first two elements of nums and delete them.The score of the operation is the sum of the deleted elements.Your task is to find the maximum number of operations that can be performed, such that all operations have the same score.Return the maximum number of operations possible that satisfy the condition mentioned above.

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class Solution { public: int minOperations(vector<int>& nums, int k) { } }; complete the function using folloiwing statement User 100231. Minimum Operations to Exceed Threshold Value I User Accepted:1 User Tried:3 Total Accepted:1 Total Submissions:3 Difficulty:Easy You are given a 0-indexed integer array nums, and an integer k. In one operation, you can remove one occurrence of the smallest element of nums. Return the minimum number of operations needed so that all elements of the array are greater than or equal to k. Example 1: Input: nums = [2,11,10,1,3], k = 10 Output: 3 Explanation: After one operation, nums becomes equal to [2, 11, 10, 3]. After two operations, nums becomes equal to [11, 10, 3]. After three operations, nums becomes equal to [11, 10]. At this stage, all the elements of nums are greater than or equal to 10 so we can stop. It can be shown that 3 is the minimum number of operations needed so that all elements of the array are greater than or equal to 10. Example 2: Input: nums = [1,1,2,4,9], k = 1 Output: 0 Explanation: All elements of the array are greater than or equal to 1 so we do not need to apply any operations on nums. Example 3: Input: nums = [1,1,2,4,9], k = 9 Output: 4 Explanation: only a single element of nums is greater than or equal to 9 so we need to apply the operations 4 times on nums. Constraints: 1 <= nums.length <= 50 1 <= nums[i] <= 109 1 <= k <= 109 The input is generated such that there is at least one index i such that nums[i] >= k.

Reversing the first K elements of a Queue

You are playing a game with your friend Ajay where Ajay gave you an array 'A' consisting of 'N' integers such that all the elements of the array does not exceed the value 'K'.Ajay has asked you to replace the minimum number of elements (replacement is the following operation: choose some index i from 1 to n and replace a[i] with some integer in range [1;k] ) to satisfy the following conditions:after all replacements, all a[i] are positive integers not greater than k;for all i from 1 to n/2 the following equation is true: a[i]+a[n]−i+1=x, where x should be the same for all n/2 pairs of elements.Note: It is guaranteed that 'N' is even.Input FormatThe first line contains two integers n and k — the length of a and the maximum possible value of some a[i] correspondingly. It is guratanteed that n is even .The second line of the test case contains n integers a1,a2,…,an, where ai is the i-th element of a.Constraints2≤n≤2⋅10^51≤k≤2⋅10^51≤a[i]≤kOutput FormatPrint the integer — the minimum number of elements you have to replace in a to satisfy the conditions from the problem statement.Sample Input 04 41 1 3 1Sample Output 01Sample Input 16 71 1 1 1 2 2Sample Output 11

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