LeetCode 3264: Final Array State After K Multiplication Operations I Solution

Master LeetCode problem 3264 (Final Array State After K Multiplication Operations I), a easy 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.

3264. Final Array State After K Multiplication Operations I

Easy

Array Math Heap (Priority Queue)Simulation

Problem Explanation

Explanation

To solve this problem, we need to perform k operations on the given array nums. In each operation, we find the minimum value x in nums and replace it with x * multiplier. We repeat this process for k operations and return the final state of nums.

The algorithmic idea involves iterating through the array nums for each operation, finding the minimum value, updating it, and continuing until k operations are completed.

Steps:

Initialize a variable minIdx to store the index of the minimum element found.
Repeat the following steps for k operations: a. Find the minimum element in nums and store its index in minIdx. b. Update the element at index minIdx with the product of the element and multiplier.
Return the updated nums array as the final state.

Time Complexity

The time complexity of this algorithm is O(k * n), where n is the length of the input array nums.

Space Complexity

The space complexity is O(1) as we are modifying the input array in-place.

Solution Code

class Solution {
    public int[] finalArrayState(int[] nums, int k, int multiplier) {
        for (int i = 0; i < k; i++) {
            int minIdx = 0;
            int minVal = Integer.MAX_VALUE;
            for (int j = 0; j < nums.length; j++) {
                if (nums[j] < minVal) {
                    minVal = nums[j];
                    minIdx = j;
                }
            }
            nums[minIdx] *= multiplier;
        }
        return nums;
    }
}

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