I have solved the second question without TLE here is my solution - 

vector<int> isNinjaAlive(int n, int k, int q, vector<int>& shield){
	
	vector<int> ans(n,k-q);

	for(int i=0;i<shield.size();i++){
		if(shield[i]>0)ans[shield[i]-1] += 1;
    }

	vector<int> res(n,0);

	for(int i=0;i<n;i++){
		if(ans[i]>0) res[i] = 1;
	}

	return res;
}

So this is how I solved B. This gives AC for 10/11 test cases and probably gives TLE for the last TC.

 int[] ninja=new int[n];
        Arrays.fill(ninja,k);
        int j=0;
        while(q-->0)
        {
           int ind=shield[j]-1;
           for(int i=0;i<n;i++)
           {
               if(i!=ind)
               {
                
                   ninja[i]--;
               }
           }
           j++;
        }
        int[] ans=new int[n];
        for(int i=0;i<n;i++)
        ans[i]=ninja[i]>0?1:0;
        return ans;

  As per the problem we need to iterate through every index of shields[] and decrease the value of every index in the ninja[] array by 1, except the index pointed to by shields[i]. Can anyone please share their AC code as how to solve this in T.C less than the T.C of my code. Was thinking of trying out a few other methods involving strings ,etc to somehow pass the time limit but couldn't implement it as I started the contest very late (with only 20 minutes remaining). Thanks.

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I was not able to pass the last test case of third question …

What i understood is that if between any 2 answers eg 001 and 110 .. if the no of different answers is divisible by 2 so they would share same rank for some or the other combination of answers…Hence if the no of different answers is odd then they would never share same rank for any combination of ans

[no of different ans = sum of instances where both answer doesnt match (eg 1 & 0 or 0 & 1) of two ninjas)]

Can anyone tell what was wrong that i passed only 10/11 test cases…or atleast share your solution

Here is my code

public class Solution {
    static long findPairs(int n, int m, int [][]response) {
  

        int cnt = 0;

        for(int i = 0;i<n;i++)
        {
           for(int j = i+1;j<n;j++)
           {
               int diff = 0;
               for(int k = 0;k<m;k++)
               {
                   if(response[i][k]!=response[j][k])diff++;
               }
               if(diff%2!=0)cnt++;
           }
        }
        return cnt;

    }
}

Hey, I was able to solve the first 2 questions !! First one was easy though but the second one was throwing TLE.  After Thinking a while i was able to reduce the Time-complexity to O(n). Slow start but yeah looking forward to participate and able to solve more and more questions. Cheers !