1769. Minimum Number of Operations to Move All Balls to Each Box

Description

You have n boxes. You are given a binary string boxes of length n, where boxes[i] is '0' if the i^th box is empty, and '1' if it contains one ball.

In one operation, you can move one ball from a box to an adjacent box. Box i is adjacent to box j if abs(i - j) == 1. Note that after doing so, there may be more than one ball in some boxes.

Return an array answer of size n, where answer[i] is the minimum number of operations needed to move all the balls to the i^th box.

Each answer[i] is calculated considering the initial state of the boxes.

Example 1:

Input: boxes = "110"
Output: [1,1,3]
Explanation: The answer for each box is as follows:
1) First box: you will have to move one ball from the second box to the first box in one operation.
2) Second box: you will have to move one ball from the first box to the second box in one operation.
3) Third box: you will have to move one ball from the first box to the third box in two operations, and move one ball from the second box to the third box in one operation.

Example 2:

Input: boxes = "001011"
Output: [11,8,5,4,3,4]

Constraints:

n == boxes.length
1 <= n <= 2000
boxes[i] is either '0' or '1'.

Solutions

Solution 1

Python Code

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class Solution:
    def minOperations(self, boxes: str) -> List[int]:
        n = len(boxes)
        left = [0] * n
        right = [0] * n
        cnt = 0
        for i in range(1, n):
            if boxes[i - 1] == '1':
                cnt += 1
            left[i] = left[i - 1] + cnt
        cnt = 0
        for i in range(n - 2, -1, -1):
            if boxes[i + 1] == '1':
                cnt += 1
            right[i] = right[i + 1] + cnt
        return [a + b for a, b in zip(left, right)]

Java Code

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class Solution {
    public int[] minOperations(String boxes) {
        int n = boxes.length();
        int[] left = new int[n];
        int[] right = new int[n];
        for (int i = 1, cnt = 0; i < n; ++i) {
            if (boxes.charAt(i - 1) == '1') {
                ++cnt;
            }
            left[i] = left[i - 1] + cnt;
        }
        for (int i = n - 2, cnt = 0; i >= 0; --i) {
            if (boxes.charAt(i + 1) == '1') {
                ++cnt;
            }
            right[i] = right[i + 1] + cnt;
        }
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            ans[i] = left[i] + right[i];
        }
        return ans;
    }
}

C++ Code

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class Solution {
public:
    vector<int> minOperations(string boxes) {
        int n = boxes.size();
        int left[n];
        int right[n];
        memset(left, 0, sizeof left);
        memset(right, 0, sizeof right);
        for (int i = 1, cnt = 0; i < n; ++i) {
            cnt += boxes[i - 1] == '1';
            left[i] = left[i - 1] + cnt;
        }
        for (int i = n - 2, cnt = 0; ~i; --i) {
            cnt += boxes[i + 1] == '1';
            right[i] = right[i + 1] + cnt;
        }
        vector<int> ans(n);
        for (int i = 0; i < n; ++i) ans[i] = left[i] + right[i];
        return ans;
    }
};

Go Code

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func minOperations(boxes string) []int {
	n := len(boxes)
	left := make([]int, n)
	right := make([]int, n)
	for i, cnt := 1, 0; i < n; i++ {
		if boxes[i-1] == '1' {
			cnt++
		}
		left[i] = left[i-1] + cnt
	}
	for i, cnt := n-2, 0; i >= 0; i-- {
		if boxes[i+1] == '1' {
			cnt++
		}
		right[i] = right[i+1] + cnt
	}
	ans := make([]int, n)
	for i := range ans {
		ans[i] = left[i] + right[i]
	}
	return ans
}

TypeScript Code

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function minOperations(boxes: string): number[] {
    const n = boxes.length;
    const left = new Array(n).fill(0);
    const right = new Array(n).fill(0);
    for (let i = 1, count = 0; i < n; i++) {
        if (boxes[i - 1] == '1') {
            count++;
        }
        left[i] = left[i - 1] + count;
    }
    for (let i = n - 2, count = 0; i >= 0; i--) {
        if (boxes[i + 1] == '1') {
            count++;
        }
        right[i] = right[i + 1] + count;
    }
    return left.map((v, i) => v + right[i]);
}

Rust Code

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impl Solution {
    pub fn min_operations(boxes: String) -> Vec<i32> {
        let s = boxes.as_bytes();
        let n = s.len();
        let mut left = vec![0; n];
        let mut right = vec![0; n];
        let mut count = 0;
        for i in 1..n {
            if s[i - 1] == b'1' {
                count += 1;
            }
            left[i] = left[i - 1] + count;
        }
        count = 0;
        for i in (0..n - 1).rev() {
            if s[i + 1] == b'1' {
                count += 1;
            }
            right[i] = right[i + 1] + count;
        }
        (0..n)
            .into_iter()
            .map(|i| left[i] + right[i])
            .collect()
    }
}

C Code

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/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* minOperations(char* boxes, int* returnSize) {
    int n = strlen(boxes);
    int* left = malloc(sizeof(int) * n);
    int* right = malloc(sizeof(int) * n);
    memset(left, 0, sizeof(int) * n);
    memset(right, 0, sizeof(int) * n);
    for (int i = 1, count = 0; i < n; i++) {
        if (boxes[i - 1] == '1') {
            count++;
        }
        left[i] = left[i - 1] + count;
    }
    for (int i = n - 2, count = 0; i >= 0; i--) {
        if (boxes[i + 1] == '1') {
            count++;
        }
        right[i] = right[i + 1] + count;
    }
    int* ans = malloc(sizeof(int) * n);
    for (int i = 0; i < n; i++) {
        ans[i] = left[i] + right[i];
    }
    free(left);
    free(right);
    *returnSize = n;
    return ans;
}

Solution 2