We find the first pair from the right where nums[i] < nums[i+1]. This i is the peak to be changed. Then find the smallest number in the right suffix that is larger than nums[i], swap them, and reverse the suffix.

1. Find i = len(nums) - 2 such that nums[i] < nums[i+1].
2. If no such i (falling sequence), reverse the whole array.
3. If i exists:
   • Find j = len(nums) - 1 such that nums[j] > nums[i].
   • Swap nums[i] and nums[j].
   • Reverse the suffix nums[i+1:].

• Time Complexity: O(N).
• Space Complexity: O(1).

def next_permutation(nums):
    n = len(nums)
    i = n - 2
    while i >= 0 and nums[i] >= nums[i + 1]:
        i -= 1
        
    if i >= 0:
        j = n - 1
        while nums[j] <= nums[i]:
            j -= 1
        nums[i], nums[j] = nums[j], nums[i]
        
    # Reverse suffix
    left, right = i + 1, n - 1
    while left < right:
        nums[left], nums[right] = nums[right], nums[left]
        left += 1
        right -= 1