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Minimum Swaps To Make Sequences Increasing in C++
Suppose we have two integer sequences A and B of the same non-zero length. We can swap elements A[i] and B[i]. We have to keep in mind that both elements are in the same index position in their respective sequences. After completing some number of swaps, A and B are both strictly increasing. We have to find the minimum number of swaps to make both sequences strictly increasing.
So if the input is like A = [1,3,5,4] and B = [1,2,3,7], then the answer will be 1, if we swap A[3] with B[3], then the sequences will be A = [1,3,5,7] and B = [1,2,3,4], both are strictly increasing.
To solve this, we will follow these steps −
n := size of A array, make two arrays swapCnt and noSwapCnt of size n each
insert 1 into swapCnt and 0 into noSwapCnt
for i in range 1 to n – 1
swapCnt[i] := n and noSwapCnt := n
if A[i] > A[i – 1] AND B[i] > B[i – 1], then
noSwapCnt[i] := noSwapCnt[i – 1]
swapCnt[i] := swapCnt[i – 1] + 1
if A[i] > B[i – 1] and B[i] > A[i – 1], then
swapCnt[i] := minimum of swapCnt[i], 1 + noSwapCnt[i – 1]
noSwapCnt[i] := minimum of swapCnt[i – 1], noSwapCnt[i]
return minimum of swapCnt[n – 1], noSwapCnt[n – 1]
Example(C++)
Let us see the following implementation to get a better understanding −
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int minSwap(vector<int>& A, vector<int>& B) {
int n = A.size();
vector <int> swapCnt(n), noSwapCnt(n);
swapCnt[0] = 1;
noSwapCnt[0] = 0;
for(int i = 1; i < n; i++){
swapCnt[i] = n;
noSwapCnt[i] = n;
if(A[i] > A[i - 1] && B[i] > B[i - 1]){
noSwapCnt[i] = noSwapCnt[i - 1];
swapCnt[i] = swapCnt[i - 1] + 1;
}
if(A[i] > B[i - 1] && B[i] > A[i - 1]){
swapCnt[i] = min(swapCnt[i], 1 + noSwapCnt[i - 1]);
noSwapCnt[i] = min(swapCnt[i - 1], noSwapCnt[i]);
}
}
return min(swapCnt[n - 1], noSwapCnt[n - 1]);
}
};
main(){
vector<int> v1 = {1,3,5,4};
vector<int> v2 = {1,2,3,7};
Solution ob;
cout << (ob.minSwap(v1, v2));
}Input
[1,3,5,4] [1,2,3,7]
Output
1
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