LeetCode 3266: Final Array State After K Multiplication Operations II Solution

Master LeetCode problem 3266 (Final Array State After K Multiplication Operations II), a hard 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.

3266. Final Array State After K Multiplication Operations II

Hard

Array 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, replace it with x * multiplier, and apply modulo 10^9 + 7 to every value in nums. We repeat this process k times and return the final state of nums after all operations.

Here is the step-by-step algorithm:

Initialize two pointers "left" and "right" to track the subarray where the modifications will occur.
While k > 0, find the minimum value in the subarray nums[left:right+1].
Update the value at nums[left] with minVal * multiplier.
Increment the left pointer to signify the modification.
Decrement k to track the number of operations remaining.
Apply modulo 10^9 + 7 to every value in nums.

Time complexity: O(k)
Space complexity: O(1)

Solution Code

class Solution {
    public int[] finalArrayState(int[] nums, int k, int multiplier) {
        int n = nums.length;
        int left = 0, right = n - 1;
        
        while(k > 0) {
            int minIndex = left;
            int minVal = nums[left];
            for(int i = left + 1; i <= right; i++) {
                if(nums[i] < minVal) {
                    minIndex = i;
                    minVal = nums[i];
                }
            }
            
            nums[minIndex] *= multiplier;
            left++;
            k--;
        }
        
        int mod = (int)1e9 + 7;
        for(int i = 0; i < n; i++) {
            nums[i] %= mod;
        }
        
        return nums;
    }
}

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