# 06.01.2025 [1769. Minimum Number of Operations to Move All Balls to Each Box]

Sum distances to all `1` for every position #medium #prefix_sum

06.01.2025

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Problem TLDR

Sum distances to all 1 for every position #medium #prefix_sum

Intuition

Let's observe an example:

    // 012345
    // 001011
    // * 2 45  2+4+5=11, right = 3
    //  *1 34  11-3=8, right = 3
    //   * 23  8-3=5 , left = 1, right = 2
    //    *12  5-2=3, +1=4
    //     *   3-2=1, +2=3, right = 1, left = 2
    //      *  1-1, 2+2=4

• the minimum operations of moving all 1 to position i is the sum of the distances

• we can reuse the previous position result: all 1's to the right became closer, and all 1's to the left increase distance, so we do sum[i + 1] = sum[i] - right_ones + left_ones

Approach

• we don't need a separate variable for the left and right, as we always operate on the balance = left - right

• careful with the operations order

• single-pass is impossible, as we should know the balance on the first position already

Complexity

• Time complexity: $$O(n)$$

• Space complexity: $$O(1)$$

Code

    fun minOperations(boxes: String): IntArray {
        var b = 0; var s = 0
        for ((i, c) in boxes.withIndex()) 
            if (c > '0') { b--; s += i }
        return IntArray(boxes.length) { i ->
            s.also { b += 2 * (boxes[i] - '0'); s += b }
        }
    }

    pub fn min_operations(boxes: String) -> Vec<i32> {
        let (mut b, mut s) = (0, 0);
        for (i, c) in boxes.bytes().enumerate() {
            if c > b'0' { b -= 1; s += i as i32 }
        }
        boxes.bytes().enumerate().map(|(i, c)| {
            let r = s; b += 2 * (c - b'0') as i32; s += b; r
        }).collect()
    }

    vector<int> minOperations(string boxes) {
        int b = 0, s = 0; vector<int> r(boxes.size());
        for (int i = 0; i < boxes.size(); ++i)
            if (boxes[i] > '0') b--, s += i;
        for (int i = 0; i < boxes.size(); ++i) 
            r[i] = s, b += 2 * (boxes[i] - '0'), s += b;
        return r;
    }