LeetCode #2835 — HARD

Minimum Operations to Form Subsequence With Target Sum

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given a 0-indexed array nums consisting of non-negative powers of 2, and an integer target.

In one operation, you must apply the following changes to the array:

Choose any element of the array nums[i] such that nums[i] > 1.
Remove nums[i] from the array.
Add two occurrences of nums[i] / 2 to the end of nums.

Return the minimum number of operations you need to perform so that nums contains a subsequence whose elements sum to target. If it is impossible to obtain such a subsequence, return -1.

A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.

Example 1:

Input: nums = [1,2,8], target = 7
Output: 1
Explanation: In the first operation, we choose element nums[2]. The array becomes equal to nums = [1,2,4,4].
At this stage, nums contains the subsequence [1,2,4] which sums up to 7.
It can be shown that there is no shorter sequence of operations that results in a subsequnce that sums up to 7.

Example 2:

Input: nums = [1,32,1,2], target = 12
Output: 2
Explanation: In the first operation, we choose element nums[1]. The array becomes equal to nums = [1,1,2,16,16].
In the second operation, we choose element nums[3]. The array becomes equal to nums = [1,1,2,16,8,8]
At this stage, nums contains the subsequence [1,1,2,8] which sums up to 12.
It can be shown that there is no shorter sequence of operations that results in a subsequence that sums up to 12.

Example 3:

Input: nums = [1,32,1], target = 35
Output: -1
Explanation: It can be shown that no sequence of operations results in a subsequence that sums up to 35.

Constraints:

1 <= nums.length <= 1000
1 <= nums[i] <= 2³⁰
nums consists only of non-negative powers of two.
1 <= target < 2³¹

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed array nums consisting of non-negative powers of 2, and an integer target. In one operation, you must apply the following changes to the array: Choose any element of the array nums[i] such that nums[i] > 1. Remove nums[i] from the array. Add two occurrences of nums[i] / 2 to the end of nums. Return the minimum number of operations you need to perform so that nums contains a subsequence whose elements sum to target. If it is impossible to obtain such a subsequence, return -1. A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.

Baseline thinking

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

Pattern signal: Array · Greedy · Bit Manipulation

Example 1

[1,2,8]
7

Example 2

[1,32,1,2]
12

Example 3

[1,32,1]
35

Core Insight

What unlocks the optimal approach

<div class="_1l1MA">if <code>target > sum(nums[i]) </code>, return <code>-1</code>. Otherwise, an answer exists</div>
<div class="_1l1MA">Solve the problem for each set bit of <code>target</code>, independently, from least significant to most significant bit. </div>
<div class="_1l1MA">For each set <code>bit</code> of <code>target</code> from least to most significant, let <code>X = sum(nums[i])</code> for <code>nums[i] <= 2^bit</code>.</div>
<div class="_1l1MA"> if <code>X >= 2^bit</code>, repeatedly select the maximum <code>nums[i]</code> such that <code>nums[i]<=2^bit</code> that has not been selected yet, until the sum of selected elements equals <code>2^bit</code>. The selected <code>nums[i]</code> will be part of the subsequence whose elements sum to target, so those elements can not be selected again. </div>
<div class="_1l1MA">Otherwise, select the smallest <code>nums[i]</code> such that <code>nums[i] > 2^bit</code>, delete <code>nums[i]</code> and add two occurences of <code>nums[i]/2</code>. Without moving to the next <code>bit</code>, go back to the step in hint 3.</div>

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

Largest constraint values

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 #2835: Minimum Operations to Form Subsequence With Target Sum
class Solution {
    public int minOperations(List<Integer> nums, int target) {
        long s = 0;
        int[] cnt = new int[32];
        for (int x : nums) {
            s += x;
            for (int i = 0; i < 32; ++i) {
                if ((x >> i & 1) == 1) {
                    ++cnt[i];
                }
            }
        }
        if (s < target) {
            return -1;
        }
        int i = 0, j = 0;
        int ans = 0;
        while (true) {
            while (i < 32 && (target >> i & 1) == 0) {
                ++i;
            }
            if (i == 32) {
                return ans;
            }
            while (j < i) {
                cnt[j + 1] += cnt[j] / 2;
                cnt[j] %= 2;
                ++j;
            }
            while (cnt[j] == 0) {
                cnt[j] = 1;
                ++j;
            }
            ans += j - i;
            --cnt[j];
            j = i;
            ++i;
        }
    }
}

// Accepted solution for LeetCode #2835: Minimum Operations to Form Subsequence With Target Sum
func minOperations(nums []int, target int) (ans int) {
	s := 0
	cnt := [32]int{}
	for _, x := range nums {
		s += x
		for i := 0; i < 32; i++ {
			if x>>i&1 > 0 {
				cnt[i]++
			}
		}
	}
	if s < target {
		return -1
	}
	var i, j int
	for {
		for i < 32 && target>>i&1 == 0 {
			i++
		}
		if i == 32 {
			return
		}
		for j < i {
			cnt[j+1] += cnt[j] >> 1
			cnt[j] %= 2
			j++
		}
		for cnt[j] == 0 {
			cnt[j] = 1
			j++
		}
		ans += j - i
		cnt[j]--
		j = i
		i++
	}
}

# Accepted solution for LeetCode #2835: Minimum Operations to Form Subsequence With Target Sum
class Solution:
    def minOperations(self, nums: List[int], target: int) -> int:
        s = sum(nums)
        if s < target:
            return -1
        cnt = [0] * 32
        for x in nums:
            for i in range(32):
                if x >> i & 1:
                    cnt[i] += 1
        i = j = 0
        ans = 0
        while 1:
            while i < 32 and (target >> i & 1) == 0:
                i += 1
            if i == 32:
                break
            while j < i:
                cnt[j + 1] += cnt[j] // 2
                cnt[j] %= 2
                j += 1
            while cnt[j] == 0:
                cnt[j] = 1
                j += 1
            ans += j - i
            cnt[j] -= 1
            j = i
            i += 1
        return ans

// Accepted solution for LeetCode #2835: Minimum Operations to Form Subsequence With Target Sum
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2835: Minimum Operations to Form Subsequence With Target Sum
// class Solution {
//     public int minOperations(List<Integer> nums, int target) {
//         long s = 0;
//         int[] cnt = new int[32];
//         for (int x : nums) {
//             s += x;
//             for (int i = 0; i < 32; ++i) {
//                 if ((x >> i & 1) == 1) {
//                     ++cnt[i];
//                 }
//             }
//         }
//         if (s < target) {
//             return -1;
//         }
//         int i = 0, j = 0;
//         int ans = 0;
//         while (true) {
//             while (i < 32 && (target >> i & 1) == 0) {
//                 ++i;
//             }
//             if (i == 32) {
//                 return ans;
//             }
//             while (j < i) {
//                 cnt[j + 1] += cnt[j] / 2;
//                 cnt[j] %= 2;
//                 ++j;
//             }
//             while (cnt[j] == 0) {
//                 cnt[j] = 1;
//                 ++j;
//             }
//             ans += j - i;
//             --cnt[j];
//             j = i;
//             ++i;
//         }
//     }
// }

// Accepted solution for LeetCode #2835: Minimum Operations to Form Subsequence With Target Sum
function minOperations(nums: number[], target: number): number {
    let s = 0;
    const cnt: number[] = Array(32).fill(0);
    for (const x of nums) {
        s += x;
        for (let i = 0; i < 32; ++i) {
            if ((x >> i) & 1) {
                ++cnt[i];
            }
        }
    }
    if (s < target) {
        return -1;
    }
    let [ans, i, j] = [0, 0, 0];
    while (1) {
        while (i < 32 && ((target >> i) & 1) === 0) {
            ++i;
        }
        if (i === 32) {
            return ans;
        }
        while (j < i) {
            cnt[j + 1] += cnt[j] >> 1;
            cnt[j] %= 2;
            ++j;
        }
        while (cnt[j] == 0) {
            cnt[j] = 1;
            j++;
        }
        ans += j - i;
        cnt[j]--;
        j = i;
        i++;
    }
}

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 M)

Space

O(log M)

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.