Next Permutation – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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

C++ Next Permutation LeetCode Solution

void nextPermutation(vector<int>& nums) {
    int i = nums.size() - 1, k = i;
    while (i > 0 && nums[i-1] >= nums[i])
        i--;
    for (int j=i; j<k; j++, k--)
        swap(nums[j], nums[k]);
    if (i > 0) {
        k = i--;
        while (nums[k] <= nums[i])
            k++;
        swap(nums[i], nums[k]);
    }
}

Java Next Permutation LeetCode Solution

public void nextPermutation(int[] A) {
    if(A == null || A.length <= 1) return;
    int i = A.length - 2;
    while(i >= 0 && A[i] >= A[i + 1]) i--; // Find 1st id i that breaks descending order
    if(i >= 0) {                           // If not entirely descending
        int j = A.length - 1;              // Start from the end
        while(A[j] <= A[i]) j--;           // Find rightmost first larger id j
        swap(A, i, j);                     // Switch i and j
    }
    reverse(A, i + 1, A.length - 1);       // Reverse the descending sequence
}

public void swap(int[] A, int i, int j) {
    int tmp = A[i];
    A[i] = A[j];
    A[j] = tmp;
}

public void reverse(int[] A, int i, int j) {
    while(i < j) swap(A, i++, j--);
}
 

Python 3 Next Permutation LeetCode Solution

def nextPermutation(self, nums):
    i = j = len(nums)-1
    while i > 0 and nums[i-1] >= nums[i]:
        i -= 1
    if i == 0:   # nums are in descending order
        nums.reverse()
        return 
    k = i - 1    # find the last "ascending" position
    while nums[j] <= nums[k]:
        j -= 1
    nums[k], nums[j] = nums[j], nums[k]  
    l, r = k+1, len(nums)-1  # reverse the second part
    while l < r:
        nums[l], nums[r] = nums[r], nums[l]
        l +=1 ; r -= 1

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Linked List-69
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Number Theory-16
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Randomized-12
Monotonic Queue-9
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Concurrency-9
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