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