LeetCode #3086 — HARD

Minimum Moves to Pick K Ones

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

The Problem

Problem Statement

You are given a binary array nums of length n, a positive integer k and a non-negative integer maxChanges.

Alice plays a game, where the goal is for Alice to pick up k ones from nums using the minimum number of moves. When the game starts, Alice picks up any index aliceIndex in the range [0, n - 1] and stands there. If nums[aliceIndex] == 1 , Alice picks up the one and nums[aliceIndex] becomes 0(this does not count as a move). After this, Alice can make any number of moves (including zero) where in each move Alice must perform exactly one of the following actions:

Select any index j != aliceIndex such that nums[j] == 0 and set nums[j] = 1. This action can be performed at most maxChanges times.
Select any two adjacent indices x and y (|x - y| == 1) such that nums[x] == 1, nums[y] == 0, then swap their values (set nums[y] = 1 and nums[x] = 0). If y == aliceIndex, Alice picks up the one after this move and nums[y] becomes 0.

Return the minimum number of moves required by Alice to pick exactly k ones.

Example 1:

Input: nums = [1,1,0,0,0,1,1,0,0,1], k = 3, maxChanges = 1

Output: 3

Explanation: Alice can pick up 3 ones in 3 moves, if Alice performs the following actions in each move when standing at aliceIndex == 1:

At the start of the game Alice picks up the one and nums[1] becomes 0. nums becomes [1,0,0,0,0,1,1,0,0,1].
Select j == 2 and perform an action of the first type. nums becomes [1,0,1,0,0,1,1,0,0,1]
Select x == 2 and y == 1, and perform an action of the second type. nums becomes [1,1,0,0,0,1,1,0,0,1]. As y == aliceIndex, Alice picks up the one and nums becomes [1,0,0,0,0,1,1,0,0,1].
Select x == 0 and y == 1, and perform an action of the second type. nums becomes [0,1,0,0,0,1,1,0,0,1]. As y == aliceIndex, Alice picks up the one and nums becomes [0,0,0,0,0,1,1,0,0,1].

Note that it may be possible for Alice to pick up 3 ones using some other sequence of 3 moves.

Example 2:

Input: nums = [0,0,0,0], k = 2, maxChanges = 3

Output: 4

Explanation: Alice can pick up 2 ones in 4 moves, if Alice performs the following actions in each move when standing at aliceIndex == 0:

Select j == 1 and perform an action of the first type. nums becomes [0,1,0,0].
Select x == 1 and y == 0, and perform an action of the second type. nums becomes [1,0,0,0]. As y == aliceIndex, Alice picks up the one and nums becomes [0,0,0,0].
Select j == 1 again and perform an action of the first type. nums becomes [0,1,0,0].
Select x == 1 and y == 0 again, and perform an action of the second type. nums becomes [1,0,0,0]. As y == aliceIndex, Alice picks up the one and nums becomes [0,0,0,0].

Constraints:

2 <= n <= 10⁵
0 <= nums[i] <= 1
1 <= k <= 10⁵
0 <= maxChanges <= 10⁵
maxChanges + sum(nums) >= k

Step 01

Brute Force Baseline

Problem summary: You are given a binary array nums of length n, a positive integer k and a non-negative integer maxChanges. Alice plays a game, where the goal is for Alice to pick up k ones from nums using the minimum number of moves. When the game starts, Alice picks up any index aliceIndex in the range [0, n - 1] and stands there. If nums[aliceIndex] == 1 , Alice picks up the one and nums[aliceIndex] becomes 0(this does not count as a move). After this, Alice can make any number of moves (including zero) where in each move Alice must perform exactly one of the following actions: Select any index j != aliceIndex such that nums[j] == 0 and set nums[j] = 1. This action can be performed at most maxChanges times. Select any two adjacent indices x and y (|x - y| == 1) such that nums[x] == 1, nums[y] == 0, then swap their values (set nums[y] = 1 and nums[x] = 0). If y == aliceIndex, Alice picks up the one

Baseline thinking

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

Pattern signal: Array · Greedy · Sliding Window

Example 1

[1,1,0,0,0,1,1,0,0,1]
3
1

Example 2

[0,0,0,0]
2
3

Core Insight

What unlocks the optimal approach

Ones created using a change require <code>2</code> moves. Hence except for the immediate neighbors of the index where we move all the ones, we should try to use change operations.
For some subset of ones, it is always better to move the ones to the median position.
We only need to be concerned with the indices where <code>nums[i] == 1</code>.

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 #3086: Minimum Moves to Pick K Ones
class Solution {
    public long minimumMoves(int[] nums, int k, int maxChanges) {
        int n = nums.length;
        int[] cnt = new int[n + 1];
        long[] s = new long[n + 1];
        for (int i = 1; i <= n; ++i) {
            cnt[i] = cnt[i - 1] + nums[i - 1];
            s[i] = s[i - 1] + i * nums[i - 1];
        }
        long ans = Long.MAX_VALUE;
        for (int i = 1; i <= n; ++i) {
            long t = 0;
            int need = k - nums[i - 1];
            for (int j = i - 1; j <= i + 1; j += 2) {
                if (need > 0 && 1 <= j && j <= n && nums[j - 1] == 1) {
                    --need;
                    ++t;
                }
            }
            int c = Math.min(need, maxChanges);
            need -= c;
            t += c * 2;
            if (need <= 0) {
                ans = Math.min(ans, t);
                continue;
            }
            int l = 2, r = Math.max(i - 1, n - i);
            while (l <= r) {
                int mid = (l + r) >> 1;
                int l1 = Math.max(1, i - mid), r1 = Math.max(0, i - 2);
                int l2 = Math.min(n + 1, i + 2), r2 = Math.min(n, i + mid);
                int c1 = cnt[r1] - cnt[l1 - 1];
                int c2 = cnt[r2] - cnt[l2 - 1];
                if (c1 + c2 >= need) {
                    long t1 = 1L * c1 * i - (s[r1] - s[l1 - 1]);
                    long t2 = s[r2] - s[l2 - 1] - 1L * c2 * i;
                    ans = Math.min(ans, t + t1 + t2);
                    r = mid - 1;
                } else {
                    l = mid + 1;
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3086: Minimum Moves to Pick K Ones
func minimumMoves(nums []int, k int, maxChanges int) int64 {
	n := len(nums)
	cnt := make([]int, n+1)
	s := make([]int, n+1)

	for i := 1; i <= n; i++ {
		cnt[i] = cnt[i-1] + nums[i-1]
		s[i] = s[i-1] + i*nums[i-1]
	}

	ans := math.MaxInt64

	for i := 1; i <= n; i++ {
		t := 0
		need := k - nums[i-1]

		for _, j := range []int{i - 1, i + 1} {
			if need > 0 && 1 <= j && j <= n && nums[j-1] == 1 {
				need--
				t++
			}
		}

		c := min(need, maxChanges)
		need -= c
		t += c * 2

		if need <= 0 {
			ans = min(ans, t)
			continue
		}

		l, r := 2, max(i-1, n-i)

		for l <= r {
			mid := (l + r) >> 1
			l1, r1 := max(1, i-mid), max(0, i-2)
			l2, r2 := min(n+1, i+2), min(n, i+mid)

			c1 := cnt[r1] - cnt[l1-1]
			c2 := cnt[r2] - cnt[l2-1]

			if c1+c2 >= need {
				t1 := c1*i - (s[r1] - s[l1-1])
				t2 := s[r2] - s[l2-1] - c2*i
				ans = min(ans, t+t1+t2)
				r = mid - 1
			} else {
				l = mid + 1
			}
		}
	}

	return int64(ans)
}

# Accepted solution for LeetCode #3086: Minimum Moves to Pick K Ones
class Solution:
    def minimumMoves(self, nums: List[int], k: int, maxChanges: int) -> int:
        n = len(nums)
        cnt = [0] * (n + 1)
        s = [0] * (n + 1)
        for i, x in enumerate(nums, 1):
            cnt[i] = cnt[i - 1] + x
            s[i] = s[i - 1] + i * x
        ans = inf
        max = lambda x, y: x if x > y else y
        min = lambda x, y: x if x < y else y
        for i, x in enumerate(nums, 1):
            t = 0
            need = k - x
            for j in (i - 1, i + 1):
                if need > 0 and 1 <= j <= n and nums[j - 1] == 1:
                    need -= 1
                    t += 1
            c = min(need, maxChanges)
            need -= c
            t += c * 2
            if need <= 0:
                ans = min(ans, t)
                continue
            l, r = 2, max(i - 1, n - i)
            while l <= r:
                mid = (l + r) >> 1
                l1, r1 = max(1, i - mid), max(0, i - 2)
                l2, r2 = min(n + 1, i + 2), min(n, i + mid)
                c1 = cnt[r1] - cnt[l1 - 1]
                c2 = cnt[r2] - cnt[l2 - 1]
                if c1 + c2 >= need:
                    t1 = c1 * i - (s[r1] - s[l1 - 1])
                    t2 = s[r2] - s[l2 - 1] - c2 * i
                    ans = min(ans, t + t1 + t2)
                    r = mid - 1
                else:
                    l = mid + 1
        return ans

// Accepted solution for LeetCode #3086: Minimum Moves to Pick K Ones
// 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 #3086: Minimum Moves to Pick K Ones
// class Solution {
//     public long minimumMoves(int[] nums, int k, int maxChanges) {
//         int n = nums.length;
//         int[] cnt = new int[n + 1];
//         long[] s = new long[n + 1];
//         for (int i = 1; i <= n; ++i) {
//             cnt[i] = cnt[i - 1] + nums[i - 1];
//             s[i] = s[i - 1] + i * nums[i - 1];
//         }
//         long ans = Long.MAX_VALUE;
//         for (int i = 1; i <= n; ++i) {
//             long t = 0;
//             int need = k - nums[i - 1];
//             for (int j = i - 1; j <= i + 1; j += 2) {
//                 if (need > 0 && 1 <= j && j <= n && nums[j - 1] == 1) {
//                     --need;
//                     ++t;
//                 }
//             }
//             int c = Math.min(need, maxChanges);
//             need -= c;
//             t += c * 2;
//             if (need <= 0) {
//                 ans = Math.min(ans, t);
//                 continue;
//             }
//             int l = 2, r = Math.max(i - 1, n - i);
//             while (l <= r) {
//                 int mid = (l + r) >> 1;
//                 int l1 = Math.max(1, i - mid), r1 = Math.max(0, i - 2);
//                 int l2 = Math.min(n + 1, i + 2), r2 = Math.min(n, i + mid);
//                 int c1 = cnt[r1] - cnt[l1 - 1];
//                 int c2 = cnt[r2] - cnt[l2 - 1];
//                 if (c1 + c2 >= need) {
//                     long t1 = 1L * c1 * i - (s[r1] - s[l1 - 1]);
//                     long t2 = s[r2] - s[l2 - 1] - 1L * c2 * i;
//                     ans = Math.min(ans, t + t1 + t2);
//                     r = mid - 1;
//                 } else {
//                     l = mid + 1;
//                 }
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3086: Minimum Moves to Pick K Ones
function minimumMoves(nums: number[], k: number, maxChanges: number): number {
    const n = nums.length;
    const cnt = Array(n + 1).fill(0);
    const s = Array(n + 1).fill(0);

    for (let i = 1; i <= n; i++) {
        cnt[i] = cnt[i - 1] + nums[i - 1];
        s[i] = s[i - 1] + i * nums[i - 1];
    }

    let ans = Infinity;
    for (let i = 1; i <= n; i++) {
        let t = 0;
        let need = k - nums[i - 1];

        for (let j of [i - 1, i + 1]) {
            if (need > 0 && 1 <= j && j <= n && nums[j - 1] === 1) {
                need--;
                t++;
            }
        }

        const c = Math.min(need, maxChanges);
        need -= c;
        t += c * 2;

        if (need <= 0) {
            ans = Math.min(ans, t);
            continue;
        }

        let l = 2,
            r = Math.max(i - 1, n - i);

        while (l <= r) {
            const mid = (l + r) >> 1;
            const [l1, r1] = [Math.max(1, i - mid), Math.max(0, i - 2)];
            const [l2, r2] = [Math.min(n + 1, i + 2), Math.min(n, i + mid)];

            const c1 = cnt[r1] - cnt[l1 - 1];
            const c2 = cnt[r2] - cnt[l2 - 1];

            if (c1 + c2 >= need) {
                const t1 = c1 * i - (s[r1] - s[l1 - 1]);
                const t2 = s[r2] - s[l2 - 1] - c2 * i;
                ans = Math.min(ans, t + t1 + t2);
                r = mid - 1;
            } else {
                l = mid + 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(n)

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.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.