Get the Shadow: Given an unsorted array A[] of size N of positive integers. One number ‘a’ from set {1, 2, N} is missing and one number ‘b’ occurs twice in array. The task is to find the repeating and the missing.
Get the Shadow Contest
Input:
The first line of input contains an integer T denoting the number of test cases. The description of T test cases follows. The first line of each test case contains a single integer N denoting the size of array. The second line contains N space-separated integers A1, A2, …, AN denoting the elements of the array.
Output:
For each testcase, in a new line, print b, which is the repeating number, followed by a, which is the missing number, in a single line.
Constraints:
1 ≤ T ≤ 100
1 ≤ N ≤ 106
1 ≤ A[i] ≤ N
Example:
Input:
2
2
2 2
3
1 3 3
Output:
2 1
3 2
Explanation:
Testcase 1: Repeating number is 2 and smallest positive missing number is 1.
Testcase 2: Repeating number is 3 and smallest positive missing number is 2.
Solution:
#include <bits/stdc++.h>
using namespace std;
void repeated_Numbers(vector<int> A) {
long long int acta = 0;
long long int actsumsq = 0;
long long int expb;
long long int expsumsq;
long long int i = 0;
for(i = 0; i < A.size(); i++){
acta = acta + (long long int)A[i];
actsumsq = actsumsq + (long long int)A[i]*A[i];
}
expb = (long long int)(A.size())*(A.size()+1)/2;
expsumsq = (long long int)(A.size())*(A.size()+1)*(2*A.size()+1)/6;
long long int diffsumsq = expsumsq - actsumsq;
long long int diffa = expb - acta;
long long int toggle = diffsumsq/diffa;
long long int miss = (toggle + diffa)/2;
long long int repp = miss - diffa;
cout<<(int)repp<<" "<<(int)miss<<endl;
return ;
}
int main() {
int T;
cin>>T;
while(T--){
int N;
cin>>N;
vector<int> vect;
for(int i=0; i<N; i++){
int temp;
cin>>temp;
vect.push_back(temp);
}
repeated_Numbers(vect);
}
return 0;
}
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