This video explains the next permutation problem which uses a very unique concept of creating the next greater sequence from an array. I have shown the entire intuition step by step using simple examples with reasons for each step.Please watch the entire video to get all the required concepts.

class Solution {
public:
    void nextPermutation(vector<int>& nums) {
        int n = nums.size();
        if(n==1)
            return;
        
        int i=1;
        int lastInc = -1;
        while(i<n) {    //Find the peak of last ASC order
            if(nums[i]>nums[i-1])
                lastInc = i;
            i+=1;
        }
        
        if(lastInc==-1) {   //Check if array is DSC
            for(i=0;i<n/2;++i)
                swap(nums[i],nums[n-i-1]);
            return;
        }
        //Find element in the range (nums[lastInc-1] to nums[lastInc]) to the right
        int mn = nums[lastInc];
        int index = lastInc;
        for(i=lastInc;i<n;++i) {
            if(nums[i]>nums[lastInc-1] and nums[i]<nums[index]) {
                index = i;
            }
        }
        swap(nums[lastInc-1],nums[index]);
        sort(nums.begin()+lastInc,nums.end());
    }
};