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19 .find non – duplicate number

Observation : a^a == 0 Code : #include<bits/stdc++.h> using namespace std; //all number will be duplicate accept one int main() { int t,n,i,x=0; cin>>t; while (t--) { int n,arr[100]; cin >> n; //n - odd number for(int i=0;i<n;i++) cin >> arr[i]; int xo = 0; for(int i=0;i<n;i++) /// O(n*n) { xo=xo^arr[i]; } cout << xo << "\n"; ...

12. toggle all bit

N = 19 = (10011) 2 Bit = 5 Bit = 1<<Bit = 10000 Bit | (Bit -1) = 10000 | 01111 = 11111 Toggle all bit = N ^ Bit = 10011 ^ 11111 = 10011 = 11111 = 01100 (all bit toggled) Code : #include<bits/stdc++.h> using namespace std; int main() { int t,n,bit,i,m; cin>>n; bit = floor(log2(n)+1...

18. swap using xor

X = 5 = 0101 Y = 9 = 1001 x = x ^ y = 0101 = 1001 = 1100 ( = 12 ) y = x ^ y = 1100 = 1001 = 0101 ( = 5 = Y) x = x ^ y = 1100 = 0101 = 1001 ( = 9 = X) Code : #include<bits/stdc++.h> using namespace std; int main() { int x,y,t; cin>>t; while (t--) { cin>>x>>y; x = x^y; y = x^y; x = x^y; cout<<x<<' '<<y<<endl; ...

11. toggle the ith bit

N = 15 , i=2 N = 1111 à 1011 (after reset the 2 nd bit) i=1<<i; = 0100 N ^ i = 1111^0100 = 1111 = 0100 = 1011 (2 nd bit from the right is toggled) Code : #include<bits/stdc++.h> using namespace std; int main() { // if i-th bit = 0 make 1 // if i-th bit = 1 make 0 int t,n,i,f,ans; cin>>t; while (t--) { cin>>n; f = 1<<i; ans = n^f; cout<<ans<<endl; ...

16 . uppercase and lowercase convert .

A -> 01000001 a -> 01100001 B -> 01000010 b -> 01100010 C -> 01000011 c -> 01100011 . . Z -> 01011010 z -> 01111010 Observations : fifth bit from the right is set from lowercase serial . Mask = 1<<5 = 00100000 Ch = Ch | Mask : uppercase -> lowercase Ch = Ch &( ~Mask ) : lowercase -> uppercase Code : int t,n,i; cin>>t; while(t--) { ...

14. check power of two .

2 0 = 1 = 000001 = 1 & (1-1) =000001 & 000000 = 000000 2 1 = 2 = 000010 = 2 & (2-1) =000010 & 000001 = 000000 2 2 = 4 = 000100 = 4 & (4-1) =000100 & 000011 = 000000 2 3 = 8 = 001000 = 8 & (8-1) =001000 & 000111 = 000000 2 4 = 16 = 010000 = 16 & (16-1) =010000 & 001111 = 000000 2 5 = 32 = 100000 = 32 & (32-1) =100000 & 011111 = 000000 Two observations : 1 . number of set bit for each number is always one 2 . n&(n-1) ...

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