LeetCode 1053: Previous Permutation With One Swap Solution
Master LeetCode problem 1053 (Previous Permutation With One Swap), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
Problem Explanation
Explanation
To find the lexicographically largest permutation that is smaller than the given array with exactly one swap, we can start from the end of the array and find the first pair of elements where the first element is greater than the second element. This is the element that we need to swap to get the desired permutation. After finding this pair, we need to find the largest element to the left of this pair that is smaller than the second element in the pair. We swap these two elements to get the final result.
- Start from the end of the array and find the first pair (i, j) where arr[i] > arr[j].
- Find the largest element k to the left of j such that arr[k] < arr[j].
- Swap arr[k] and arr[j] to get the final result.
Time Complexity: O(n) where n is the number of elements in the array. Space Complexity: O(1)
Solution Code
class Solution {
public int[] prevPermOpt1(int[] arr) {
int i = arr.length - 2;
while (i >= 0 && arr[i] <= arr[i + 1]) {
i--;
}
if (i >= 0) {
int j = arr.length - 1;
while (arr[j] >= arr[i] || arr[j] == arr[j - 1]) {
j--;
}
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
return arr;
}
}Try It Yourself
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