Yes

I also had a stupid bug. After fixing my code gives these graphs

• k=1 There are 4 diameters (1,3), (3,5), (4,5), (2,5)
• k=2 There are 5 diameters (1,3,5), (3,5,4), (3,5,4,1), (3,5,2), (2,5,4)
• k=3 There are 3 diameters (1,2,3,4,5), (1,3,5,4), (1,3,5,2)
• k=4 There are 0 diameters

what is wrong in it?

static int f(int left,int right,int last,char a[]){
        if(dp[left][right][last]!=null)return dp[left][right][last];
        if(left>=right)return dp[left][right][last]=0;
        if(a[left]!=a[right]){
            return dp[left][right][last]=Math.max(f(left+1,right,last,a),f(left,right-1,last,a));
        }
        else{
            if(last!=a[left]-'a'){
                int temp=2+f(left+1,right-1,a[left]-'a',a);
                temp=Math.max(temp,Math.max(f(left+1,right,last,a),f(left,right-1,last,a)));
                return dp[left][right][last]=temp;
            }
            else{
                return dp[left][right][last]=Math.max(f(left+1,right,last,a),f(left,right-1,last,a));
            }
        }
    }

That wasn't exactly "Google-able", you had to build up the stirling number recurrence anyways and not everyone remembers the expansion of ith level coefficients of nth stirling number and there is nothing wrong with googling it.
Our opinions might differ, ofcourse.

Great.

I remember there was a check box before starting the test on HackerRank that said only language reference and IDE with auto-completion are allowed?

I did the cubes one, and I passed 12 test cases/15 in the metrices problem. What was your approach for War?