LeetCode #3776 — MEDIUM

Minimum Moves to Balance Circular Array

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

The Problem

Problem Statement

You are given a circular array balance of length n, where balance[i] is the net balance of person i.

In one move, a person can transfer exactly 1 unit of balance to either their left or right neighbor.

Return the minimum number of moves required so that every person has a non-negative balance. If it is impossible, return -1.

Note: You are guaranteed that at most 1 index has a negative balance initially.

Example 1:

Input: balance = [5,1,-4]

Output: 4

Explanation:

One optimal sequence of moves is:

Move 1 unit from i = 1 to i = 2, resulting in balance = [5, 0, -3]
Move 1 unit from i = 0 to i = 2, resulting in balance = [4, 0, -2]
Move 1 unit from i = 0 to i = 2, resulting in balance = [3, 0, -1]
Move 1 unit from i = 0 to i = 2, resulting in balance = [2, 0, 0]

Thus, the minimum number of moves required is 4.

Example 2:

Input: balance = [1,2,-5,2]

Output: 6

Explanation:

One optimal sequence of moves is:

Move 1 unit from i = 1 to i = 2, resulting in balance = [1, 1, -4, 2]
Move 1 unit from i = 1 to i = 2, resulting in balance = [1, 0, -3, 2]
Move 1 unit from i = 3 to i = 2, resulting in balance = [1, 0, -2, 1]
Move 1 unit from i = 3 to i = 2, resulting in balance = [1, 0, -1, 0]
Move 1 unit from i = 0 to i = 1, resulting in balance = [0, 1, -1, 0]
Move 1 unit from i = 1 to i = 2, resulting in balance = [0, 0, 0, 0]

Thus, the minimum number of moves required is 6.

Example 3:

Input: balance = [-3,2]

Output: -1

Explanation:

It is impossible to make all balances non-negative for balance = [-3, 2], so the answer is -1.

Constraints:

1 <= n == balance.length <= 10⁵
-10⁹ <= balance[i] <= 10⁹
There is at most one negative value in balance initially.

Step 01

Brute Force Baseline

Problem summary: You are given a circular array balance of length n, where balance[i] is the net balance of person i. In one move, a person can transfer exactly 1 unit of balance to either their left or right neighbor. Return the minimum number of moves required so that every person has a non-negative balance. If it is impossible, return -1. Note: You are guaranteed that at most 1 index has a negative balance initially.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

[5,1,-4]

Example 2

[1,2,-5,2]

Example 3

[-3,2]

Step 02

Core Insight

What unlocks the optimal approach

If there is no negative value, then the answer is 0. If the total sum is less than 0, then the answer is -1.
Sort the positive values in <code>nums</code> by their distance from the index with the negative value.
Greedily use as many values as needed from the sorted <code>nums</code> to offset the current negative value.

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 #3776: Minimum Moves to Balance Circular Array
class Solution {
    public long minMoves(int[] balance) {
        long sum = 0;
        for (int b : balance) {
            sum += b;
        }
        if (sum < 0) {
            return -1;
        }

        int n = balance.length;
        int mn = balance[0];
        int idx = 0;
        for (int i = 1; i < n; i++) {
            if (balance[i] < mn) {
                mn = balance[i];
                idx = i;
            }
        }

        if (mn >= 0) {
            return 0;
        }

        int need = -mn;
        long ans = 0;

        for (int j = 1; j < n; j++) {
            int a = balance[(idx - j + n) % n];
            int b = balance[(idx + j) % n];

            int c1 = Math.min(a, need);
            need -= c1;
            ans += (long) c1 * j;

            int c2 = Math.min(b, need);
            need -= c2;
            ans += (long) c2 * j;
        }

        return ans;
    }
}

// Accepted solution for LeetCode #3776: Minimum Moves to Balance Circular Array
func minMoves(balance []int) int64 {
	var sum int64
	for _, b := range balance {
		sum += int64(b)
	}
	if sum < 0 {
		return -1
	}

	n := len(balance)
	mn := balance[0]
	idx := 0
	for i := 1; i < n; i++ {
		if balance[i] < mn {
			mn = balance[i]
			idx = i
		}
	}

	if mn >= 0 {
		return 0
	}

	need := -mn
	var ans int64

	for j := 1; j < n; j++ {
		a := balance[(idx-j+n)%n]
		b := balance[(idx+j)%n]

		c1 := min(a, need)
		need -= c1
		ans += int64(c1) * int64(j)

		c2 := min(b, need)
		need -= c2
		ans += int64(c2) * int64(j)
	}

	return ans
}

# Accepted solution for LeetCode #3776: Minimum Moves to Balance Circular Array
class Solution:
    def minMoves(self, balance: List[int]) -> int:
        if sum(balance) < 0:
            return -1
        mn = min(balance)
        if mn >= 0:
            return 0
        need = -mn
        i = balance.index(mn)
        n = len(balance)
        ans = 0
        for j in range(1, n):
            a = balance[(i - j + n) % n]
            b = balance[(i + j - n) % n]
            c1 = min(a, need)
            need -= c1
            ans += c1 * j
            c2 = min(b, need)
            need -= c2
            ans += c2 * j
        return ans

// Accepted solution for LeetCode #3776: Minimum Moves to Balance Circular Array
// 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 #3776: Minimum Moves to Balance Circular Array
// class Solution {
//     public long minMoves(int[] balance) {
//         long sum = 0;
//         for (int b : balance) {
//             sum += b;
//         }
//         if (sum < 0) {
//             return -1;
//         }
// 
//         int n = balance.length;
//         int mn = balance[0];
//         int idx = 0;
//         for (int i = 1; i < n; i++) {
//             if (balance[i] < mn) {
//                 mn = balance[i];
//                 idx = i;
//             }
//         }
// 
//         if (mn >= 0) {
//             return 0;
//         }
// 
//         int need = -mn;
//         long ans = 0;
// 
//         for (int j = 1; j < n; j++) {
//             int a = balance[(idx - j + n) % n];
//             int b = balance[(idx + j) % n];
// 
//             int c1 = Math.min(a, need);
//             need -= c1;
//             ans += (long) c1 * j;
// 
//             int c2 = Math.min(b, need);
//             need -= c2;
//             ans += (long) c2 * j;
//         }
// 
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3776: Minimum Moves to Balance Circular Array
function minMoves(balance: number[]): number {
    const sum = balance.reduce((a, b) => a + b, 0);
    if (sum < 0) {
        return -1;
    }

    const n = balance.length;
    let mn = balance[0],
        idx = 0;
    for (let i = 1; i < n; i++) {
        if (balance[i] < mn) {
            mn = balance[i];
            idx = i;
        }
    }

    if (mn >= 0) {
        return 0;
    }

    let need = -mn;
    let ans = 0;

    for (let j = 1; j < n; j++) {
        const a = balance[(idx - j + n) % n];
        const b = balance[(idx + j) % n];

        const c1 = Math.min(a, need);
        need -= c1;
        ans += c1 * j;

        const c2 = Math.min(b, need);
        need -= c2;
        ans += c2 * j;
    }

    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.