Can you think of extending the logic for a normal stone paper scissor game to get the solution?

What if we check for a solution by first retracting the left hand and then the right hand? Then each checking becomes same as checking for normal stone paper scissor game.

To solve the problem, follow these steps:

Key Observations

1. If Ritesh retracts his left hand (L1), his remaining hand (L2) will compete against Aaryan's hands (L3 and L4).

2. If Ritesh retracts his right hand (L2), his remaining hand (L1) will compete against Aaryan's hands (L3 and L4).

3. For each scenario, compare Ritesh's remaining hand with both of Aaryan's hands:

• Ritesh wins if his symbol beats Aaryan's symbol.
• Ritesh loses if Aaryan's symbol beats his.
• The result is a draw if both symbols are the same.

Decision Process

• If Ritesh can avoid losing against both of Aaryan's hands by retracting either hand, output Left.
• If Ritesh can avoid losing by retracting only one specific hand, output Left or Right, depending on the safe choice.
• If both scenarios result in at least one loss, output Impossible.

#include <iostream>
#include <string>
using namespace std;
string determineOutcome(char r1, char a1) {
    if (r1 == a1) {
        return "Draw";
    } else if ((r1 == 'R' && a1 == 'S') || (r1 == 'S' && a1 == 'P') || (r1 == 'P' && a1 == 'R')) {
        return "Win";
    } else {
        return "Lose";
    }
}
int main() {
    int t;
    cin >> t;
    while (t--) {
        string s;
        cin >> s;
        char rLeft = s[0];
        char rRight = s[1];
        char aLeft = s[2];
        char aRight = s[3];
        string outcomeLeft1 = determineOutcome(rRight, aLeft);
        string outcomeLeft2 = determineOutcome(rRight, aRight);
        string outcomeRight1 = determineOutcome(rLeft, aLeft);
        string outcomeRight2 = determineOutcome(rLeft, aRight);
        bool leftSafe = (outcomeLeft1 != "Lose" && outcomeLeft2 != "Lose");
        bool rightSafe = (outcomeRight1 != "Lose" && outcomeRight2 != "Lose");
        if (leftSafe && rightSafe) {
            cout << "Left" << endl;
        } else if (leftSafe) {
            cout << "Left" << endl;
        } else if (rightSafe) {
            cout << "Right" << endl;
        } else {
            cout << "Impossible" << endl;
        }
    }
    return 0;
}

We need to convert all ones to zeroes .

After each operation, what changes happen to the 1's present in the string? Try writing out a few examples in binary and observe the pattern .

Look carefully at how $$$X$$$ $$$\text&$$$ $$$(-X)$$$ affects different positions .

Can any 1 except the rightmost one be changed by applying one operation?

Only the right-most 1 is converted to 0 after applying one operation. So it is easy to say that the answer to the question is simply the number of 1's in the binary representation .

But why only the right most 1 ??

Let $$$X$$$ be any binary string , for example —

$$$X$$$ = $$$1101.....10000$$$

then $$$(-X)$$$ will be calculated as follows —

Flipping the bits , we get — $$$0010.....01111$$$

Now adding 1 we get — $$$0010.....10000$$$

So $$$X = 1101.....10000$$$ and $$$(-X) = 0010.....10000$$$ , we can now easily see that all the bits on the left side of the right most 1 are opposite in $$$X$$$ and $$$(-X)$$$ , so $$$ X \text& (-X)$$$ will be equal to $$$0000.....10000$$$ and subtracting it will convert the rightmost $$$1$$$ to $$$0$$$ .

#include <bits/stdc++.h>
 
using namespace std ;
 
#define int long long 
#define endl '\n'
 
void solve(){
    
    string s ;
    cin >> s ;
    
    int n = s.size() ;
    
    int ans = 0 ;
    
    for(int i=0 ; i<n ; i++){
        if(s[i]=='1'){
            ans++ ;
        }
    }
    
    cout<<ans<<endl;
}
 
signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
	
	int t = 1 ;
	cin >> t ;
	
	while(t--){
	    solve();
	}
}

Consider inserting a value $$$x$$$ between two numbers $$$a$$$ and $$$b$$$.

Original contribution to score: $$$|a-b|$$$

New contribution after insertion: $$$|a-x| + |x-b|$$$

For score to remain same: $$$|a-b| = |a-x| + |x-b|$$$

This is the triangle inequality!

For each possible position $$$i$$$, we need numbers that work between $$$a[i]$$$ and $$$a[i+1]$$$ This gives us ranges $$$[\min(a_i,a_{i+1}), \max(a_i,a_{i+1})]$$$ for each position.

Now we just need to find the union of all such intervals. But do we actually need to manually calculate the union ? What observation can be made about the union of all such intervals ?

One observation that we can make is that the union of ranges $$$[\min(a_i,a_{i+1}), \max(a_i,a_{i+1})]$$$ and $$$[\min(a_{i+1},a_{i+2}), \max(a_{i+1},a_{i+2})]$$$ is simply $$$[\min(a_i,a_{i+1},a_{i+2}), \max(a_i,a_{i+1},a_{i+2})]$$$

Similary we can say that union of ranges , $$$[\min(a_i,a_{i+1}), \max(a_i,a_{i+1})]$$$ , $$$[\min(a_{i+1},a_{i+2}), \max(a_{i+1},a_{i+2})]$$$ and $$$[\min(a_{i+2},a_{i+3}), \max(a_{i+2},a_{i+3})]$$$ is $$$[\min(a_i,a_{i+1},a_{i+2},a_{i+3}), \max(a_i,a_{i+1},a_{i+2},a_{i+3})]$$$ and so on .

So for the complete union , we get —

Required Interval = $$$[\min(a_i,a_{i+1},...,a_n) , \max(a_i,a_{i+1},...,a_n)]$$$

And Number of elements in this interval = MaxElement — MinElement + 1

#include <bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
 
using namespace std;
using namespace __gnu_pbds;
 
#define int long long 
#define endl '\n'
 
 
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds; // *find_by_order, order_of_key
 
 
void solve(){
    
    int n ;
    cin >> n ;
 
    vector<int> arr(n-1) ;
    
    for(int i=0 ; i<n-1 ; i++){
        cin >> arr[i] ;
    }
    
    int mx = *max_element(arr.begin(), arr.end()) ;
    int mn = *min_element(arr.begin(), arr.end()) ;
    
    cout<<mx+1-mn<<endl;
    
}
 
signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
	
	int t = 1 ;
	cin >> t ;
	
	while(t--){
	    solve();
	}
}

Can the parity p ever change, irrespective of how we choose $$$a$$$ and $$$b$$$?

When is the initial parity p odd?

1. Initial Sum Parity:

• The sum of the first $$$n$$$ natural numbers is given by:

• For $$$S$$$ to be odd, $$$n(n+1)$$$ must be divisible by $$$2$$$ but not by $$$4$$$. This happens when $$$n$$$ leaves a remainder of $$$1$$$ or $$$2$$$ when divided by $$$4$$$.

1. Parity Preservation:

• During each operation, the sum $$$S$$$ changes as:

• This is same as decreasing $$$S$$$ by $$$2 * $$$ min($$$a,b$$$), which will always be even. Thus if $$$S$$$ was initially odd, then odd — even = odd. Thus, the parity of the sum remains unchanged after each operation.

To solve the problem, we need to check if the initial sum $$$S$$$ is odd. This happens if and only if $$$n$$$ leaves a remainder of $$$1$$$ or $$$2$$$ when divided by $$$4$$$. For each test case, we can compute $$$n$$$ mod $$$4$$$ and check if the result is $$$1$$$ or $$$2$$$.

#include <bits/stdc++.h>
using namespace std;
#define int long long
signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        if (n % 4 == 1 || n % 4 == 2)
            cout << "YES" << endl;
        else
            cout << "NO" << endl;
    }
    return 0;
}