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1053. Previous Permutation With One Swap

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

• Java
• C++
• Python
• Go
• TypeScript

• 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;
      }
  };
  

• 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
  
  

• 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;
  }
  
  

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