• + 2 comments

    No need for that young fella 🧐

    JAVA SOLUTION

    100% TEST CASES ✅

    COMPLEXITY IN 0(N) TIME

        static String fibonacciModified(int t1, int t2, int n) {
            BigInteger[] arr = new BigInteger[n];
            BigInteger tOne = new BigInteger(Integer.toString(t1));
            BigInteger tTwo = new BigInteger(Integer.toString(t2));
            arr[0] = tOne;
            arr[1] = tTwo;
            for(int i = 2; i < n; i++){
                BigInteger iMinus1BigInteger = arr[i - 1];
                BigInteger iMinus2BigInteger = arr[i - 2];
                BigInteger powered = iMinus1BigInteger.pow(2);
                BigInteger poweredAndSum = powered.add(iMinus2BigInteger);
                arr[i] = poweredAndSum;
            }
            BigInteger result = arr[n-1];
            return result.toString();
        }
    
    • + 0 comments

      but we have to return int value not string, so how we can do that

    • + 0 comments

      ''' static BigInteger fibonacciModified(int t1, int t2, int n) { BigInteger[] arr = new BigInteger[n]; BigInteger tOne = new BigInteger(Integer.toString(t1)); BigInteger tTwo = new BigInteger(Integer.toString(t2)); arr[0] = tOne; arr[1] = tTwo; for(int i = 2; i < n; i++){ BigInteger iMinus1BigInteger = arr[i - 1]; BigInteger iMinus2BigInteger = arr[i - 2]; BigInteger powered = iMinus1BigInteger.pow(2); BigInteger poweredAndSum = powered.add(iMinus2BigInteger); arr[i] = poweredAndSum; }''' BigInteger result = arr[n-1];

          return result;
      
      }