2948. Make Lexicographically Smallest Array by Swapping Elements

Description

You are given a 0-indexed array of positive integers nums and a positive integer limit.

In one operation, you can choose any two indices i and j and swap nums[i] and nums[j] if |nums[i] - nums[j]| <= limit.

Return the lexicographically smallest array that can be obtained by performing the operation any number of times.

An array a is lexicographically smaller than an array b if in the first position where a and b differ, array a has an element that is less than the corresponding element in b. For example, the array [2,10,3] is lexicographically smaller than the array [10,2,3] because they differ at index 0 and 2 < 10.

Example 1:

Input: nums = [1,5,3,9,8], limit = 2
Output: [1,3,5,8,9]
Explanation: Apply the operation 2 times:
- Swap nums[1] with nums[2]. The array becomes [1,3,5,9,8]
- Swap nums[3] with nums[4]. The array becomes [1,3,5,8,9]
We cannot obtain a lexicographically smaller array by applying any more operations.
Note that it may be possible to get the same result by doing different operations.

Example 2:

Input: nums = [1,7,6,18,2,1], limit = 3
Output: [1,6,7,18,1,2]
Explanation: Apply the operation 3 times:
- Swap nums[1] with nums[2]. The array becomes [1,6,7,18,2,1]
- Swap nums[0] with nums[4]. The array becomes [2,6,7,18,1,1]
- Swap nums[0] with nums[5]. The array becomes [1,6,7,18,1,2]
We cannot obtain a lexicographically smaller array by applying any more operations.

Example 3:

Input: nums = [1,7,28,19,10], limit = 3
Output: [1,7,28,19,10]
Explanation: [1,7,28,19,10] is the lexicographically smallest array we can obtain because we cannot apply the operation on any two indices.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹
1 <= limit <= 10⁹

Solutions

Solution 1: Sorting

According to the problem description, the sorted array \(\textit{nums}\) can be partitioned into several subarrays such that the difference between adjacent elements in each subarray does not exceed \(\textit{limit}\).

The lexicographically smallest array obtainable through swapping is thus the one where the elements within each subarray are sorted and placed back into their original positions in order.

We first pair each element in \(\textit{nums}\) with its index to form an array of tuples, then sort by element value. We then traverse the sorted tuple array, identify the range of each subarray, sort the elements within each subarray by their original indices, and fill them back into the corresponding positions to obtain the final result.

The time complexity is \(O(n \times \log n)\) and the space complexity is \(O(n)\), where \(n\) is the length of the array \(\textit{nums}\).

Python3JavaC++GoTypeScriptRust

class Solution:
    def lexicographicallySmallestArray(self, nums: List[int], limit: int) -> List[int]:
        n = len(nums)
        arr = sorted(zip(nums, range(n)))
        ans = [0] * n
        i = 0
        while i < n:
            j = i + 1
            while j < n and arr[j][0] - arr[j - 1][0] <= limit:
                j += 1
            idx = sorted(k for _, k in arr[i:j])
            for k, (x, _) in zip(idx, arr[i:j]):
                ans[k] = x
            i = j
        return ans

class Solution {
    public int[] lexicographicallySmallestArray(int[] nums, int limit) {
        int n = nums.length;
        Integer[] idx = new Integer[n];
        Arrays.setAll(idx, i -> i);
        Arrays.sort(idx, (i, j) -> nums[i] - nums[j]);
        int[] ans = new int[n];
        for (int i = 0; i < n;) {
            int j = i + 1;
            while (j < n && nums[idx[j]] - nums[idx[j - 1]] <= limit) {
                ++j;
            }
            Integer[] t = Arrays.copyOfRange(idx, i, j);
            Arrays.sort(t, (x, y) -> x - y);
            for (int k = i; k < j; ++k) {
                ans[t[k - i]] = nums[idx[k]];
            }
            i = j;
        }
        return ans;
    }
}

class Solution {
public:
    vector<int> lexicographicallySmallestArray(vector<int>& nums, int limit) {
        int n = nums.size();
        vector<int> idx(n);
        iota(idx.begin(), idx.end(), 0);
        sort(idx.begin(), idx.end(), [&](int i, int j) {
            return nums[i] < nums[j];
        });
        vector<int> ans(n);
        for (int i = 0; i < n;) {
            int j = i + 1;
            while (j < n && nums[idx[j]] - nums[idx[j - 1]] <= limit) {
                ++j;
            }
            vector<int> t(idx.begin() + i, idx.begin() + j);
            sort(t.begin(), t.end());
            for (int k = i; k < j; ++k) {
                ans[t[k - i]] = nums[idx[k]];
            }
            i = j;
        }
        return ans;
    }
};

func lexicographicallySmallestArray(nums []int, limit int) []int {
    n := len(nums)
    idx := make([]int, n)
    for i := range idx {
        idx[i] = i
    }
    slices.SortFunc(idx, func(i, j int) int { return nums[i] - nums[j] })
    ans := make([]int, n)
    for i := 0; i < n; {
        j := i + 1
        for j < n && nums[idx[j]]-nums[idx[j-1]] <= limit {
            j++
        }
        t := slices.Clone(idx[i:j])
        slices.Sort(t)
        for k := i; k < j; k++ {
            ans[t[k-i]] = nums[idx[k]]
        }
        i = j
    }
    return ans
}

function lexicographicallySmallestArray(nums: number[], limit: number): number[] {
    const n: number = nums.length;
    const idx: number[] = Array.from({ length: n }, (_, i) => i);
    idx.sort((i, j) => nums[i] - nums[j]);
    const ans: number[] = Array(n).fill(0);
    for (let i = 0; i < n; ) {
        let j = i + 1;
        while (j < n && nums[idx[j]] - nums[idx[j - 1]] <= limit) {
            j++;
        }
        const t: number[] = idx.slice(i, j).sort((a, b) => a - b);
        for (let k: number = i; k < j; k++) {
            ans[t[k - i]] = nums[idx[k]];
        }
        i = j;
    }
    return ans;
}

impl Solution {
    pub fn lexicographically_smallest_array(nums: Vec<i32>, limit: i32) -> Vec<i32> {
        let n = nums.len();
        let mut idx: Vec<usize> = (0..n).collect();

        idx.sort_by_key(|&i| nums[i]);

        let mut ans = vec![0; n];

        let mut i = 0;
        while i < n {
            let mut j = i + 1;
            while j < n && nums[idx[j]] - nums[idx[j - 1]] <= limit {
                j += 1;
            }

            let mut t = idx[i..j].to_vec();
            t.sort();

            for k in i..j {
                ans[t[k - i]] = nums[idx[k]];
            }

            i = j;
        }

        ans
    }
}

2948. Make Lexicographically Smallest Array by Swapping Elements

Description

Solutions

Solution 1: Sorting

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