LeetCode #1558 — MEDIUM

Minimum Numbers of Function Calls to Make Target Array

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an integer array nums. You have an integer array arr of the same length with all values set to 0 initially. You also have the following modify function:

You want to use the modify function to convert arr to nums using the minimum number of calls.

Return the minimum number of function calls to make nums from arr.

The test cases are generated so that the answer fits in a 32-bit signed integer.

Example 1:

Input: nums = [1,5]
Output: 5
Explanation: Increment by 1 (second element): [0, 0] to get [0, 1] (1 operation).
Double all the elements: [0, 1] -> [0, 2] -> [0, 4] (2 operations).
Increment by 1 (both elements)  [0, 4] -> [1, 4] -> [1, 5] (2 operations).
Total of operations: 1 + 2 + 2 = 5.

Example 2:

Input: nums = [2,2]
Output: 3
Explanation: Increment by 1 (both elements) [0, 0] -> [0, 1] -> [1, 1] (2 operations).
Double all the elements: [1, 1] -> [2, 2] (1 operation).
Total of operations: 2 + 1 = 3.

Example 3:

Input: nums = [4,2,5]
Output: 6
Explanation: (initial)[0,0,0] -> [1,0,0] -> [1,0,1] -> [2,0,2] -> [2,1,2] -> [4,2,4] -> [4,2,5](nums).

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums. You have an integer array arr of the same length with all values set to 0 initially. You also have the following modify function: You want to use the modify function to convert arr to nums using the minimum number of calls. Return the minimum number of function calls to make nums from arr. The test cases are generated so that the answer fits in a 32-bit signed integer.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy · Bit Manipulation

Example 1

[1,5]

Example 2

[2,2]

Example 3

[4,2,5]

Step 02

Core Insight

What unlocks the optimal approach

Work backwards: try to go from nums to arr.
You should try to divide by 2 as much as possible, but you can only divide by 2 if everything is even.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1558: Minimum Numbers of Function Calls to Make Target Array
class Solution {
    public int minOperations(int[] nums) {
        int ans = 0;
        int mx = 0;
        for (int v : nums) {
            mx = Math.max(mx, v);
            ans += Integer.bitCount(v);
        }
        ans += Integer.toBinaryString(mx).length() - 1;
        return ans;
    }
}

// Accepted solution for LeetCode #1558: Minimum Numbers of Function Calls to Make Target Array
func minOperations(nums []int) int {
	ans, mx := 0, 0
	for _, v := range nums {
		mx = max(mx, v)
		for v > 0 {
			ans += v & 1
			v >>= 1
		}
	}
	if mx > 0 {
		for mx > 0 {
			ans++
			mx >>= 1
		}
		ans--
	}
	return ans
}

# Accepted solution for LeetCode #1558: Minimum Numbers of Function Calls to Make Target Array
class Solution:
    def minOperations(self, nums: List[int]) -> int:
        return sum(v.bit_count() for v in nums) + max(0, max(nums).bit_length() - 1)

// Accepted solution for LeetCode #1558: Minimum Numbers of Function Calls to Make Target Array
struct Solution;

impl Solution {
    fn min_operations(nums: Vec<i32>) -> i32 {
        let mut ones = 0;
        let mut max_width = 0;
        for mut x in nums {
            ones += x.count_ones();
            let mut width = 0;
            while x > 0 {
                x >>= 1;
                width += 1;
            }
            max_width = max_width.max(width);
        }
        ((ones + max_width) as i32 - 1).max(0)
    }
}

#[test]
fn test() {
    let nums = vec![1, 5];
    let res = 5;
    assert_eq!(Solution::min_operations(nums), res);
    let nums = vec![2, 2];
    let res = 3;
    assert_eq!(Solution::min_operations(nums), res);
    let nums = vec![4, 2, 5];
    let res = 6;
    assert_eq!(Solution::min_operations(nums), res);
    let nums = vec![3, 2, 2, 4];
    let res = 7;
    assert_eq!(Solution::min_operations(nums), res);
    let nums = vec![2, 4, 8, 16];
    let res = 8;
    assert_eq!(Solution::min_operations(nums), res);
    let nums = vec![0];
    let res = 0;
    assert_eq!(Solution::min_operations(nums), res);
}

// Accepted solution for LeetCode #1558: Minimum Numbers of Function Calls to Make Target Array
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1558: Minimum Numbers of Function Calls to Make Target Array
// class Solution {
//     public int minOperations(int[] nums) {
//         int ans = 0;
//         int mx = 0;
//         for (int v : nums) {
//             mx = Math.max(mx, v);
//             ans += Integer.bitCount(v);
//         }
//         ans += Integer.toBinaryString(mx).length() - 1;
//         return ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.