Question

1053. Previous Permutation With One Swap

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Description

Given an array of positive integers arr (not necessarily distinct), return _the __lexicographically__ largest permutation that is smaller than_ arr, that can be made with exactly one swap. If it cannot be done, then return the same array.

Note that a _swap_ exchanges the positions of two numbers arr[i] and arr[j]

 

Example 1:

Input: arr = [3,2,1]
Output: [3,1,2]
Explanation: Swapping 2 and 1.

Example 2:

Input: arr = [1,1,5]
Output: [1,1,5]
Explanation: This is already the smallest permutation.

Example 3:

Input: arr = [1,9,4,6,7]
Output: [1,7,4,6,9]
Explanation: Swapping 9 and 7.

 

Constraints:

• 1 <= arr.length <= 104
• 1 <= arr[i] <= 104

Solutions

Solution 1

class Solution:
  def prevPermOpt1(self, arr: List[int]) -> List[int]:
    n = len(arr)
    for i in range(n - 1, 0, -1):
      if arr[i - 1] > arr[i]:
        for j in range(n - 1, i - 1, -1):
          if arr[j] < arr[i - 1] and arr[j] != arr[j - 1]:
            arr[i - 1], arr[j] = arr[j], arr[i - 1]
            return arr
    return arr

class Solution {
  public int[] prevPermOpt1(int[] arr) {
    int n = arr.length;
    for (int i = n - 1; i > 0; --i) {
      if (arr[i - 1] > arr[i]) {
        for (int j = n - 1; j > i - 1; --j) {
          if (arr[j] < arr[i - 1] && arr[j] != arr[j - 1]) {
            int t = arr[i - 1];
            arr[i - 1] = arr[j];
            arr[j] = t;
            return arr;
          }
        }
      }
    }
    return arr;
  }
}

class Solution {
public:
  vector<int> prevPermOpt1(vector<int>& arr) {
    int n = arr.size();
    for (int i = n - 1; i > 0; --i) {
      if (arr[i - 1] > arr[i]) {
        for (int j = n - 1; j > i - 1; --j) {
          if (arr[j] < arr[i - 1] && arr[j] != arr[j - 1]) {
            swap(arr[i - 1], arr[j]);
            return arr;
          }
        }
      }
    }
    return arr;
  }
};

func prevPermOpt1(arr []int) []int {
  n := len(arr)
  for i := n - 1; i > 0; i-- {
    if arr[i-1] > arr[i] {
      for j := n - 1; j > i-1; j-- {
        if arr[j] < arr[i-1] && arr[j] != arr[j-1] {
          arr[i-1], arr[j] = arr[j], arr[i-1]
          return arr
        }
      }
    }
  }
  return arr
}

function prevPermOpt1(arr: number[]): number[] {
  const n = arr.length;
  for (let i = n - 1; i > 0; --i) {
    if (arr[i - 1] > arr[i]) {
      for (let j = n - 1; j > i - 1; --j) {
        if (arr[j] < arr[i - 1] && arr[j] !== arr[j - 1]) {
          const t = arr[i - 1];
          arr[i - 1] = arr[j];
          arr[j] = t;
          return arr;
        }
      }
    }
  }
  return arr;
}