72 lines
2.2 KiB
C++
72 lines
2.2 KiB
C++
//c++ program to find the next greatest permutation of given number.
|
|
//Sample input 1 :
|
|
//nums =[1,2,3]
|
|
//sample output 1:
|
|
// nums =[1,3,2]
|
|
//sample input 2:
|
|
//nums =[3,2,1]
|
|
//sample output:
|
|
//nums =[1,2,3]
|
|
//time complexity:O(n).
|
|
//space complexity:O(1),because we are doing in place.
|
|
#include<bits/stdc++.h>
|
|
using namespace std;
|
|
vector<int>nextPermutation(int n,vector<int>nums) {
|
|
|
|
int k;
|
|
int change=-1;
|
|
//Base case if vector have size either 0 or 1 element.
|
|
if(nums.size()==0 || nums.size()==1)
|
|
return nums;
|
|
//Base Case if vector have size 2,simply reverse them suppose we have nums=[1,2] then output will be [2,1].
|
|
if(nums.size()==2)
|
|
{
|
|
reverse(nums.begin(),nums.end());
|
|
|
|
}
|
|
//case where the size of vector is greater the 2 start traversing from the right side
|
|
// While going from right to left, if we find an index whose value is less than index at right
|
|
// changes in permutation starts from that index.
|
|
for(k=nums.size()-2;k>=0;k--)
|
|
{
|
|
if(nums[k]<nums[k+1])
|
|
|
|
{
|
|
//swap(nums[k],nums[k+1]);
|
|
change=k;
|
|
break;
|
|
}
|
|
|
|
}
|
|
if (change==-1)
|
|
{
|
|
reverse(nums.begin(),nums.end());
|
|
//return;
|
|
}
|
|
//Next thing that we can observe is the value at nums[change] is always replaced by the value
|
|
// which is greater than value at change index and occurring first while traversing from back.
|
|
for(k=nums.size()-1;k>change;k--)
|
|
{
|
|
if(nums[k]-nums[change]>0)
|
|
{
|
|
swap(nums[k],nums[change]);
|
|
break;
|
|
}
|
|
|
|
}
|
|
//After swapping we are just sort the values from change+1 index to n-1 index.
|
|
reverse(nums.begin()+change+1,nums.end());
|
|
return nums;
|
|
}
|
|
int main()
|
|
{
|
|
int n = 6;
|
|
vector<int>v{5,3,4,9,7,6};
|
|
vector<int> res;
|
|
res = nextPermutation(n, v);
|
|
for (int i = 0; i < res.size(); i++) {
|
|
cout << res[i] << " ";
|
|
}
|
|
}
|
|
|