Question

31. Next Permutation

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Description

A permutation of an array of integers is an arrangement of its members into a sequence or linear order.

• For example, for arr = [1,2,3], the following are all the permutations of arr: [1,2,3], [1,3,2], [2, 1, 3], [2, 3, 1], [3,1,2], [3,2,1].

The next permutation of an array of integers is the next lexicographically greater permutation of its integer. More formally, if all the permutations of the array are sorted in one container according to their lexicographical order, then the next permutation of that array is the permutation that follows it in the sorted container. If such arrangement is not possible, the array must be rearranged as the lowest possible order (i.e., sorted in ascending order).

• For example, the next permutation of arr = [1,2,3] is [1,3,2].
• Similarly, the next permutation of arr = [2,3,1] is [3,1,2].
• While the next permutation of arr = [3,2,1] is [1,2,3] because [3,2,1] does not have a lexicographical larger rearrangement.

Given an array of integers nums, _find the next permutation of_ nums.

The replacement must be in place and use only constant extra memory.

 

Example 1:

Input: nums = [1,2,3]
Output: [1,3,2]

Example 2:

Input: nums = [3,2,1]
Output: [1,2,3]

Example 3:

Input: nums = [1,1,5]
Output: [1,5,1]

 

Constraints:

• 1 <= nums.length <= 100
• 0 <= nums[i] <= 100

Solutions

Solution 1: Two traversals

We first traverse the array from back to front and find the first position $i$ where $nums[i] \lt nums[i + 1]$.

Then traverse the array from back to front again and find the first position $j$ where $nums[j] \gt nums[i]$. Swap $nums[i]$ and $nums[j]$, and then reverse the elements from $nums[i + 1]$ to $nums[n - 1]$, the next permutation can be obtained.

The time complexity is $O(n)$ and the space complexity is $O(1)$. Where $n$ is the length of the array.

class Solution:
  def nextPermutation(self, nums: List[int]) -> None:
    n = len(nums)
    i = next((i for i in range(n - 2, -1, -1) if nums[i] < nums[i + 1]), -1)
    if ~i:
      j = next((j for j in range(n - 1, i, -1) if nums[j] > nums[i]))
      nums[i], nums[j] = nums[j], nums[i]
    nums[i + 1 :] = nums[i + 1 :][::-1]

class Solution {
  public void nextPermutation(int[] nums) {
    int n = nums.length;
    int i = n - 2;
    for (; i >= 0; --i) {
      if (nums[i] < nums[i + 1]) {
        break;
      }
    }
    if (i >= 0) {
      for (int j = n - 1; j > i; --j) {
        if (nums[j] > nums[i]) {
          swap(nums, i, j);
          break;
        }
      }
    }

    for (int j = i + 1, k = n - 1; j < k; ++j, --k) {
      swap(nums, j, k);
    }
  }

  private void swap(int[] nums, int i, int j) {
    int t = nums[j];
    nums[j] = nums[i];
    nums[i] = t;
  }
}

class Solution {
public:
  void nextPermutation(vector<int>& nums) {
    int n = nums.size();
    int i = n - 2;
    while (~i && nums[i] >= nums[i + 1]) {
      --i;
    }
    if (~i) {
      for (int j = n - 1; j > i; --j) {
        if (nums[j] > nums[i]) {
          swap(nums[i], nums[j]);
          break;
        }
      }
    }
    reverse(nums.begin() + i + 1, nums.end());
  }
};

func nextPermutation(nums []int) {
  n := len(nums)
  i := n - 2
  for ; i >= 0 && nums[i] >= nums[i+1]; i-- {
  }
  if i >= 0 {
    for j := n - 1; j > i; j-- {
      if nums[j] > nums[i] {
        nums[i], nums[j] = nums[j], nums[i]
        break
      }
    }
  }
  for j, k := i+1, n-1; j < k; j, k = j+1, k-1 {
    nums[j], nums[k] = nums[k], nums[j]
  }
}

function nextPermutation(nums: number[]): void {
  const n = nums.length;
  let i = n - 2;
  while (i >= 0 && nums[i] >= nums[i + 1]) {
    --i;
  }
  if (i >= 0) {
    for (let j = n - 1; j > i; --j) {
      if (nums[j] > nums[i]) {
        [nums[i], nums[j]] = [nums[j], nums[i]];
        break;
      }
    }
  }
  for (let j = n - 1; j > i; --j, ++i) {
    [nums[i + 1], nums[j]] = [nums[j], nums[i + 1]];
  }
}

/**
 * @param {number[]} nums
 * @return {void} Do not return anything, modify nums in-place instead.
 */
var nextPermutation = function (nums) {
  const n = nums.length;
  let i = n - 2;
  while (i >= 0 && nums[i] >= nums[i + 1]) {
    --i;
  }
  if (i >= 0) {
    let j = n - 1;
    while (j > i && nums[j] <= nums[i]) {
      --j;
    }
    [nums[i], nums[j]] = [nums[j], nums[i]];
  }
  for (i = i + 1, j = n - 1; i < j; ++i, --j) {
    [nums[i], nums[j]] = [nums[j], nums[i]];
  }
};

public class Solution {
  public void NextPermutation(int[] nums) {
    int n = nums.Length;
    int i = n - 2;
    while (i >= 0 && nums[i] >= nums[i + 1]) {
      --i;
    }
    if (i >= 0) {
      for (int j = n - 1; j > i; --j) {
        if (nums[j] > nums[i]) {
          swap(nums, i, j);
          break;
        }
      }
    }
    for (int j = i + 1, k = n - 1; j < k; ++j, --k) {
      swap(nums, j, k);
    }
  }

  private void swap(int[] nums, int i, int j) {
    int t = nums[j];
    nums[j] = nums[i];
    nums[i] = t;
  }
}