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Leetcode 3151. Special Array I

Problem Statement

You are given a non-negative integer array numList. A special integer num is defined as the number of i (where 0 <= i < numList.length) such that numList[i] % num == 0 and numList[i] / num is not an integer. You are to return num.

Clarifying Questions

1. Can numList have repeated numbers?
   • Yes, there can be repeated numbers.
2. What should be the return value if there is no such num that satisfies the conditions described?
   • If no such num exists, we can return -1.

Strategy

We need to find a special integer num such that:

1. For all i in the range [0, numList.length), numList[i] % num == 0.
2. For the same i, numList[i] / num is not an integer.

Steps:

1. Iterate through the possible values of num from 1 to the largest number present in numList.
2. For each candidate, check the two conditions.
3. If a candidate satisfies the conditions, return it.
4. If no candidates satisfy the conditions, return -1.

Code

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

int findSpecialInteger(const std::vector<int>& numList) {
    int maxElement = *std::max_element(numList.begin(), numList.end());
    
    for (int num = 1; num <= maxElement; ++num) {
        bool isSpecial = true;
        
        for (int i = 0; i < numList.size(); ++i) {
            if (numList[i] % num != 0 || numList[i] / num == 0 || numList[i] / num == 1) {
                isSpecial = false;
                break;
            }
        }
        
        if (isSpecial) {
            return num;
        }
    }
    
    return -1;
}

int main() {
    std::vector<int> numList = {8, 12, 10, 20};
    int result = findSpecialInteger(numList);
    std::cout << "Special integer: " << result << std::endl;
    
    return 0;
}

Time Complexity

• Finding the Maximum Element: (O(n)) where (n) is the length of numList.
• Checking Each Candidate num:
  • We iterate through each candidate from 1 to the maximum element which is (O(\max(numList))).
  • For each candidate, we check all elements of numList, which is (O(n)).

Therefore, the overall time complexity is (O(n \times \max(numList))).

This solution may not be the most efficient for very large values of numList, but it should work within reasonable constraints of typical interview problems. For more optimizations, additional mathematical insights or constraints need to be evaluated.

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