LeetCode 3124: Find Longest Calls Solution

Master LeetCode problem 3124 (Find Longest Calls), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3124. Find Longest Calls

Medium

Database

Problem Explanation

Explanation:

To solve this problem, we can use dynamic programming with a bottom-up approach. We will iterate through the given array of calls and maintain two arrays to store the maximum profit we can get if we choose to make or skip a call at each position. At each step, we update these arrays based on the previous values.

Here are the steps for our algorithm:

Initialize two arrays, dpMake and dpSkip, both of size n (the number of calls).
Set dpMake[0] as the profit of the first call and dpSkip[0] as 0 (since we skip the first call).
Iterate from index 1 to n-1:
- Update dpMake[i] as the maximum profit between the profit of the current call plus dpSkip[i-1] and the profit of the previous call.
- Update dpSkip[i] as the maximum profit between the previous call's profit and dpSkip[i-1].
Return the maximum between the last element of dpMake and dpSkip.

Time Complexity: O(n) Space Complexity: O(n)

: :

Solution Code

class Solution {
    public int findLongestCalls(int[] calls) {
        int n = calls.length;
        int[] dpMake = new int[n];
        int[] dpSkip = new int[n];
        
        dpMake[0] = calls[0];
        dpSkip[0] = 0;
        
        for (int i = 1; i < n; i++) {
            dpMake[i] = Math.max(calls[i] + dpSkip[i-1], dpMake[i-1]);
            dpSkip[i] = Math.max(dpMake[i-1], dpSkip[i-1]);
        }
        
        return Math.max(dpMake[n-1], dpSkip[n-1]);
    }
}

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