leetcode/lcci/16.16.Sub Sort

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16.16. Sub Sort

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Description

Given an array of integers, write a method to find indices m and n such that if you sorted elements m through n, the entire array would be sorted. Minimize n - m (that is, find the smallest such sequence).

Return [m,n]. If there are no such m and n (e.g. the array is already sorted), return [-1, -1].

Example:

Input:  [1,2,4,7,10,11,7,12,6,7,16,18,19]
Output:  [3,9]

Note:

• 0 <= len(array) <= 1000000

Solutions

Solution 1: Two Passes

We first traverse the array array from left to right, and use mx to record the maximum value encountered so far. If the current value x is less than mx, it means that x needs to be sorted, and we record the index i of x as right; otherwise, update mx.

Similarly, we traverse the array array from right to left, and use mi to record the minimum value encountered so far. If the current value x is greater than mi, it means that x needs to be sorted, and we record the index i of x as left; otherwise, update mi.

Finally, return [left, right].

The time complexity is O(n), where n is the length of the array array. The space complexity is O(1).

Python3

class Solution:
    def subSort(self, array: List[int]) -> List[int]:
        n = len(array)
        mi, mx = inf, -inf
        left = right = -1
        for i, x in enumerate(array):
            if x < mx:
                right = i
            else:
                mx = x
        for i in range(n - 1, -1, -1):
            if array[i] > mi:
                left = i
            else:
                mi = array[i]
        return [left, right]

Java

class Solution {
    public int[] subSort(int[] array) {
        int n = array.length;
        int mi = Integer.MAX_VALUE, mx = Integer.MIN_VALUE;
        int left = -1, right = -1;
        for (int i = 0; i < n; ++i) {
            if (array[i] < mx) {
                right = i;
            } else {
                mx = array[i];
            }
        }
        for (int i = n - 1; i >= 0; --i) {
            if (array[i] > mi) {
                left = i;
            } else {
                mi = array[i];
            }
        }
        return new int[] {left, right};
    }
}

C++

class Solution {
public:
    vector<int> subSort(vector<int>& array) {
        int n = array.size();
        int mi = INT_MAX, mx = INT_MIN;
        int left = -1, right = -1;
        for (int i = 0; i < n; ++i) {
            if (array[i] < mx) {
                right = i;
            } else {
                mx = array[i];
            }
        }
        for (int i = n - 1; ~i; --i) {
            if (array[i] > mi) {
                left = i;
            } else {
                mi = array[i];
            }
        }
        return {left, right};
    }
};

Go

func subSort(array []int) []int {
	n := len(array)
	mi, mx := math.MaxInt32, math.MinInt32
	left, right := -1, -1
	for i, x := range array {
		if x < mx {
			right = i
		} else {
			mx = x
		}
	}
	for i := n - 1; i >= 0; i-- {
		if array[i] > mi {
			left = i
		} else {
			mi = array[i]
		}
	}
	return []int{left, right}
}

TypeScript

function subSort(array: number[]): number[] {
    const n = array.length;
    let [mi, mx] = [Infinity, -Infinity];
    let [left, right] = [-1, -1];
    for (let i = 0; i < n; ++i) {
        if (array[i] < mx) {
            right = i;
        } else {
            mx = array[i];
        }
    }
    for (let i = n - 1; ~i; --i) {
        if (array[i] > mi) {
            left = i;
        } else {
            mi = array[i];
        }
    }
    return [left, right];
}

Swift

class Solution {
    func subSort(_ array: [Int]) -> [Int] {
        let n = array.count
        var mi = Int.max, mx = Int.min
        var left = -1, right = -1

        for i in 0..<n {
            if array[i] < mx {
                right = i
            } else {
                mx = array[i]
            }
        }

        for i in stride(from: n - 1, through: 0, by: -1) {
            if array[i] > mi {
                left = i
            } else {
                mi = array[i]
            }
        }

        return [left, right]
    }
}