LeetCode 1558: Minimum Numbers of Function Calls to Make Target Array Solution

Master LeetCode problem 1558 (Minimum Numbers of Function Calls to Make Target Array), 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.

1558. Minimum Numbers of Function Calls to Make Target Array

Medium

Array Greedy Bit Manipulation

Problem Explanation

Explanation

To minimize the number of function calls, we need to consider two types of operations: incrementing all elements by 1 and doubling all elements. To reach the target array nums, we can first increment each element to its closest power of 2 and then use the double operation to get to the target array. The key observation is that incrementing a number by 1 is equivalent to adding 1 to its binary representation. Thus, we can count the number of set bits in each number and use this information to minimize the number of operations.

Algorithmic Idea:

Count the number of set bits in each number in the input array nums.
The total number of operations required to increment a number to its closest power of 2 is equal to the number of set bits in that number.
The total number of operations required to reach the target array from the initial array is the sum of the number of set bits in each number in nums plus the maximum number of set bits in any number in nums.

Time Complexity:

The time complexity for this approach is O(n * log(max(nums)) + n), where n is the length of the input array nums and max(nums) is the maximum number in nums.

Space Complexity:

The space complexity is O(1) as we are not using any extra space other than a few variables.

Solution Code

class Solution {
    public int minOperations(int[] nums) {
        int maxBits = 0;
        int totalOperations = 0;
        
        for (int num : nums) {
            int bits = 0;
            while (num > 0) {
                if (num % 2 == 1) {
                    totalOperations++;
                    bits++;
                }
                num /= 2;
            }
            maxBits = Math.max(maxBits, bits);
        }
        
        return totalOperations + maxBits;
    }
}

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