LeetCode Entry

1658. Minimum Operations to Reduce X to Zero

23.09.2026 medium 2026 kotlin rust

Min operations to remove first or last elements sum of x

1658. Minimum Operations to Reduce X to Zero medium substack youtube

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Problem TLDR

Min operations to remove first or last elements sum of x

Intuition

Invert the problem: longest subarray with sum equal to sum()-x

Approach

• use the target itself as a sum variable, compare with 0

Complexity

• Time complexity: \(O(n)\)

• Space complexity: \(O(1)\)

Code

    fun minOperations(n: IntArray, x: Int) = n.run {
        var t = sum() - x; var j = 0
        indices.maxOf { i ->
            t -= n[i]; while (t < 0 && j <= i) t += n[j++]
            if (t == 0) i - j + 1 else -1
        }.let { if (it < 0) -1 else size - it }
    }

    pub fn min_operations(n: Vec<i32>, x: i32) -> i32 {
        let (mut t, mut j, l) = (n.iter().sum::<i32>() - x, 0, n.len());
        (0..l).filter_map(|i| {
            t -= n[i]; while t < 0 && j <= i { t += n[j]; j += 1 }
            (t == 0).then_some((l + j - i - 1) as i32)
        }).min().unwrap_or(-1)
    }