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Leetcode 3038. Maximum Number of Operations With the Same Score I

Problem Statement

1. Maximum Number of Operations With the Same Score I-out

Given an array nums of length n, your goal is to find the maximum number of operations that can be performed such that every two chosen elements sum to the same score. An operation is defined as choosing two elements of the array and creating their sum.

Clarifying Questions

1. Clarify Input Constraints:
   • What is the range of values for elements in the array nums?
   • Is the array guaranteed to have at least two elements?
2. Output Format:
   • Should we return only the count of such operations, or do we need to return the pairs as well?
3. Sum Definition:
   • Do we need to consider all possible sums, or are there any specific sums that we need to prioritize?

Strategy

1. Understanding the Problem:
   • Given array nums, we need to find the maximum number of pairs (i, j) such that their sum nums[i] + nums[j] is the same and return the count of these operations.
2. Using a Hash Map:
   • Use a hash map to store the frequencies of sums.
   • Iterate over all pairs (i, j) in nums and calculate their sum.
   • Update the hash map with the frequency of each sum.
   • The maximum value in the hash map gives us the maximum number of operations.
3. Edge Cases:
   • Handle cases where the array length is less than 2 by returning 0.

Solution Code

#include <iostream>
#include <vector>
#include <unordered_map>
#include <algorithm>

using namespace std;

int maxOperationsWithSameScore(vector<int>& nums) {
    // Edge case: less than two elements in the array
    if (nums.size() < 2) return 0;
    
    unordered_map<int, int> sumFrequency;
    int maxOperations = 0;

    for (size_t i = 0; i < nums.size(); ++i) {
        for (size_t j = i + 1; j < nums.size(); ++j) {
            int sum = nums[i] + nums[j];
            sumFrequency[sum]++;
            maxOperations = max(maxOperations, sumFrequency[sum]);
        }
    }

    return maxOperations;
}

int main() {
    vector<int> nums = {1, 2, 3, 4, 5};
    cout << "Maximum number of operations: " << maxOperationsWithSameScore(nums) << endl;
    return 0;
}

Time Complexity

• Time Complexity: The nested loops result in O(n^2) time complexity, where n is the number of elements in the array nums.
• Space Complexity: The space complexity is O(n^2) in the worst case for the hash map storing the sums (though typically it will be much smaller).

Explanation

• Hash Map Approach: We use a nested loop to generate sums of all pairs and store their counts in a hash map.
• Updating Maximum: Track the maximum frequency of any sum during the process to get the required result.

This approach ensures checking each pair efficiently, and using a hash map helps in counting frequencies seamlessly.

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