Next Permutation of Array in DSA Python

Mental Model

Find the next bigger arrangement of numbers by changing the smallest possible part of the array.

Analogy: Imagine you have a lock with digits. To get the next code, you try to increase the last digit. If it can't go higher, you move left to find a digit you can increase, then reset digits to the right to the smallest order.

Array: [1, 2, 3, 4, 5]
Indexes:  0  1  2  3  4
We want to find the next bigger order by changing digits from right to left.

Dry Run Walkthrough

Input: array: [1, 2, 3]

Goal: Find the next permutation which is just bigger than [1, 2, 3]

Step 1: Find the first index from right where array[i] < array[i+1], here i=1 (2 < 3)

[1, 2, 3] ↑i=1

Why: We need to find where the order breaks from the right to find the pivot to increase

Step 2: Find the first index from right where array[j] > array[i], here j=2 (3 > 2)

[1, 2, 3] ↑j=2

Why: We find the number just bigger than array[i] to swap and get next bigger permutation

Step 3: Swap array[i] and array[j], array becomes [1, 3, 2]

[1, 3, 2]

Why: Swapping increases the number at i to the next bigger digit

Step 4: Reverse the subarray after i (index 2 to end), here only one element so no change

[1, 3, 2]

Why: Reversing makes the tail the smallest possible order to get the next permutation

Result:

[1, 3, 2]

Annotated Code

DSA Python

from typing import List

class Solution:
    def nextPermutation(self, nums: List[int]) -> None:
        n = len(nums)
        i = n - 2
        # Find first decreasing element from right
        while i >= 0 and nums[i] >= nums[i + 1]:
            i -= 1
        if i >= 0:
            j = n - 1
            # Find element just bigger than nums[i]
            while nums[j] <= nums[i]:
                j -= 1
            nums[i], nums[j] = nums[j], nums[i]  # Swap
        # Reverse the tail
        left, right = i + 1, n - 1
        while left < right:
            nums[left], nums[right] = nums[right], nums[left]
            left += 1
            right -= 1

if __name__ == '__main__':
    arr = [1, 2, 3]
    Solution().nextPermutation(arr)
    print(arr)

while i >= 0 and nums[i] >= nums[i + 1]: i -= 1

Find the first index from right where order breaks (nums[i] < nums[i+1])

if i >= 0:

Check if such index exists, meaning next permutation is possible

while nums[j] <= nums[i]: j -= 1

Find the element just bigger than nums[i] from right

nums[i], nums[j] = nums[j], nums[i]

Swap to increase the number at i

while left < right: nums[left], nums[right] = nums[right], nums[left] left += 1 right -= 1

Reverse the subarray after i to get the smallest tail order

Output

[1, 3, 2]

Complexity Analysis

Time: O(n) because we scan the array from right to left a few times but only once fully

Space: O(1) because we modify the array in place without extra space

vs Alternative: Naive approach would generate all permutations and find the next one, which is O(n!) and very inefficient

Edge Cases

array is in descending order like [3, 2, 1]

No next bigger permutation exists, so array is reversed to smallest order [1, 2, 3]

DSA Python

while i >= 0 and nums[i] >= nums[i + 1]:
    i -= 1

array has only one element like [1]

No change since only one permutation exists

DSA Python

while i >= 0 and nums[i] >= nums[i + 1]:
    i -= 1

array has repeated elements like [1, 5, 5]

Next permutation is found correctly by swapping and reversing tail

DSA Python

while nums[j] <= nums[i]:
    j -= 1

When to Use This Pattern

When you need to find the next bigger arrangement of numbers or characters in order, use the next permutation pattern because it efficiently finds the next lexicographical order.

Common Mistakes

Mistake: Not reversing the tail after swapping, leading to wrong next permutation
Fix: Always reverse the subarray after the swapped index to get the smallest tail order

Mistake: Swapping the wrong elements or not finding the correct pivot index
Fix: Carefully find the first decreasing element from right and then find the just bigger element to swap

Summary

Finds the next lexicographically bigger permutation of the array in place.

Use when you want to generate permutations in order or find the next bigger number with same digits.

The key is to find the pivot from right where order breaks, swap with just bigger element, then reverse the tail.