time complexity -O(n),space complexity -O(1)*/
| 1 | /*time complexity -O(n),space complexity -O(1)*/ |
| 2 | class Solution { |
| 3 | public: |
| 4 | void nextPermutation(vector<int>& nums) { |
| 5 | int n=nums.size(); |
| 6 | int j=n-1; |
| 7 | /*starting from the right side check upto which |
| 8 | index the array is non-decreasing*/ |
| 9 | |
| 10 | while(j!=0&&nums[j]<=nums[j-1]) |
| 11 | j--; |
| 12 | /* the array is in descending order which represents |
| 13 | the maximum possible lexicographical permutation*/ |
| 14 | if(j==0) |
| 15 | { |
| 16 | reverse(nums.begin(),nums.end()); |
| 17 | return; |
| 18 | } |
| 19 | j--; |
| 20 | /*nums[j] will be swapped with the smallest number |
| 21 | greater than nums[j] on the right side*/ |
| 22 | for(int i=n-1;i>j;i--) |
| 23 | { |
| 24 | if(nums[i]>nums[j]) |
| 25 | { |
| 26 | swap(nums[i],nums[j]); |
| 27 | /* new nums[j] has greater value so, sort or just |
| 28 | reverse the array from j+1 th index to last*/ |
| 29 | reverse(nums.begin()+j+1,nums.end()); |
| 30 | return; |
| 31 | } |
| 32 | } |
| 33 | } |
| 34 | }; |
nothing calls this directly
no outgoing calls
no test coverage detected