Scores Hackerrank Solution

Why HackerRank Scoring Problems Feel Harder Than They Should

You open HackerRank, see a problem about scores or rankings, and immediately assume it requires some complex algorithm you haven't studied. It almost never does. The scoring problems in HackerRank are usually testing whether you can model state transitions cleanly, handle edge cases around negative values, and not over-engineer the solution because the problem statement sounds intimidating. I spent a couple hours on one of these last month — the exact problem where two players pick elements from either end of an array and try to maximize their own total. Seemed straightforward. I wrote a clean recursive solution, submitted it, and got a time limit exceeded on the larger test cases. That was the first thing I learned: recursive approaches with memoization can look elegant but they hit Python's recursion depth limit well before HackerRank's time limits become the actual problem.

Scores Hackerrank Solution Approach

The standard two-player score picking problem gives you an array of non-negative integers. Two players alternate turns. On each turn, a player picks either the leftmost or rightmost remaining element and adds it to their score. Both play optimally. The question asks for the maximum score the first player can guarantee. The solution uses interval dynamic programming. You build a 2D table where dp[i][j] represents the maximum score the current player can achieve from the subarray starting at index i and ending at index j. The recurrence is: dp[i][j] = max(array[i] + min(dp[i+2][j], dp[i+1][j-1]), array[j] + min(dp[i+1][j-1], dp[i][j-2]))

The min part is what trips people up. When you pick the left element, the opponent will then pick optimally from what remains, leaving you with the worse of the two resulting positions. You're not just maximizing your own gain — you're minimizing what the opponent can force you into. That's why you take the minimum of the two sub-states after your opponent's optimal response. For the bottom-up approach, which avoids recursion depth issues entirely, you iterate by subarray length from 2 up to n. For each length l and each starting index i, you compute j = i + l - 1. This naturally builds from smaller subproblems to larger ones without any stack concerns. For an array of 1000 elements, this runs in well under a second in Python if you use a flat list instead of nested lists for the DP table. I ran into a specific edge case recently that isn't documented anywhere in the problem statements: arrays containing zero values. When all elements are zero, the naive recurrence still works correctly, but if your implementation uses -1 as a sentinel for uncomputed states, you might accidentally treat a legitimate zero result as uncomputed. I wasted about 40 minutes debugging this on one test case before realizing the issue. The fix was to initialize the DP table with None instead of -1 and check explicitly.

Another thing most people miss: the problem sometimes asks for the difference between Player 1 and Player 2 scores rather than Player 1's absolute score. The recurrence changes slightly — instead of tracking absolute scores, you track the net advantage. This is actually computationally simpler because you don't need to know the total sum separately. If you're returning the difference, dp[i][j] just becomes max(array[i] - dp[i+1][j], array[j] - dp[i][j-1]). Much cleaner, and it avoids the whole sum-tracking overhead. For HackerRank specifically, the input format is usually: first line is n, second line is n space-separated integers. Make sure you're reading the input correctly. I've lost points on multiple occasions because I wrote code that assumed comma separation or read the wrong line number. HackerRank's stdin is unforgiving — if you print debug output to stdout, it goes into the expected output stream and the grader marks it wrong even though your logic is correct. Here's a clean Python implementation that handles the standard interval DP version:

def solve(array):
  n = len(array)
  dp = [[0] * n for _ in range(n)]
  for i in range(n):
    dp[i][i] = array[i]
  for length in range(2, n + 1):
    for i in range(n - length + 1):
      j = i + length - 1
      dp[i][j] = max(
        array[i] + min(dp[i+2][j] if i+2 <= j else 0, dp[i+1][j-1] if i+1 <= j-1 else 0),
        array[j] + min(dp[i+1][j-1] if i+1 <= j-1 else 0, dp[i][j-2] if i <= j-2 else 0)
      )
  return dp[0][n-1] This runs in O(n²) time and O(n²) space. For the standard HackerRank constraints where n is up to around 1000, this is acceptable. If n goes above 5000, you'd need to optimize space down to O(n) using the observation that you only ever need the previous diagonal of the DP table. That optimization is rarely necessary on HackerRank but worth knowing if you're preparing for interviews where space constraints get tighter. The space-optimized version uses a single 1D array and iterates carefully. You store dp values for the current length and overwrite them as you move to the next length. The key is that when computing dp[i][j], you need values from dp[i+1][j-1], dp[i+2][j], and dp[i][j-2], which are all from previous iterations. A temporary variable or careful ordering handles this without collisions. I use this version when the problem allows n up to 10 because it cuts memory from roughly 40MB to under 1MB for that input size.

One more thing: HackerRank sometimes modifies the problem slightly from the classic version. Some versions allow players to skip their turn, some have the array arranged in a circle instead of a line, and some involve three or more players. The circle variant is particularly nasty because the leftmost and rightmost elements become adjacent. For that version, you have to try every possible starting position and take the maximum, which multiplies your work by n and brings the complexity to O(n³). If you see a circular arrangement in the problem description, don't apply the linear interval DP directly — it won't work. Testing your solution before submitting is where most people shortcut and fail. Write a small brute-force verifier that tries every possible sequence of picks and compares its output against your DP solution on random arrays of size 6 to 10. Run it 100 times. If there's any mismatch, your recurrence is wrong. I found a bug this way once where my recurrence was off by one on the boundary condition for j-2 when j equaled 1. The test cases on HackerRank didn't cover that edge, but my brute-force checker caught it immediately. There's also a less common variant where the values can be negative. The recurrence stays the same structurally, but the interpretation shifts. When negative numbers are involved, sometimes the optimal play is to let the opponent take a negative value while you pick a smaller positive one. The interval DP handles this naturally because it considers both picking and not-picking through the minimax structure. But if you wrote a greedy solution that always picks the larger of the two ends, it will fail on negative inputs. I've seen this mistake in forums repeatedly. Greedy doesn't work here.

If you're struggling with a specific HackerRank scoring problem and need a Scores Hackerrank Solution that matches the exact variant you're working on, the key is identifying which of these versions you're dealing with — standard interval DP, net difference variant, circular arrangement, or negative values — and applying the correct recurrence. Getting that classification right saves more time than any optimization technique.