Maximum Fruits Harvested After at Most K Steps LeetCode solution in Java

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class Solution {
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  public int maxTotalFruits(int[][] fruits, int startPos, int k) {
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    int n = fruits.length;
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    int posOfLastFruit = fruits[n - 1][0];
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    int prefixArr[] = new int[posOfLastFruit + 1];
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    int start = Math.max(startPos - k, 0);
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    int end = Math.min(startPos + k, prefixArr.length - 1);
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    if (startPos > posOfLastFruit) {
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      int diff = startPos - posOfLastFruit;
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      startPos = posOfLastFruit;
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      k = k - diff;
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      if (k == 0) return fruits[posOfLastFruit][1];
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      else if (k < 0) return 0;
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    }
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    for (int i = 0; i < n; i++) {
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      prefixArr[fruits[i][0]] = fruits[i][1];
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    }
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    int curr = 0;
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    for (int i = startPos - 1; i >= start; i--) {
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      curr += prefixArr[i];
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      prefixArr[i] = curr;
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    }
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    curr = 0;
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    for (int i = startPos + 1; i <= end; i++) {
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      curr += prefixArr[i];
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      prefixArr[i] = curr;
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    }
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    int minimum = prefixArr[startPos];
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    prefixArr[startPos] = 0;
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    int ans = 0;
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    for (int i = start; i < startPos; i++) {
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      int maxCurrPossible = prefixArr[i];
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      int stepsAlreadyWalked = startPos - i;
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      int stepsRemaining = k - stepsAlreadyWalked;
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      int endIndex = i + stepsRemaining;
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      if (endIndex > startPos && endIndex < prefixArr.length) {
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        maxCurrPossible += prefixArr[endIndex];
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      } else if (endIndex >= prefixArr.length) {
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        maxCurrPossible += prefixArr[prefixArr.length - 1];
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      }
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      ans = Math.max(ans, maxCurrPossible);
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    }
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    for (int i = startPos + 1; i <= end; i++) {
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      int maxCurrPossible = prefixArr[i];
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      int stepsAlreadyWalked = i - startPos;
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      int stepsRemaining = k - stepsAlreadyWalked;
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      int endIndex = i - stepsRemaining;
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      if (endIndex < startPos && endIndex >= 0) {
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        maxCurrPossible += prefixArr[endIndex];
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      } else if (endIndex < 0) {
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        maxCurrPossible += prefixArr[0];
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      }
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      ans = Math.max(ans, maxCurrPossible);
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    }
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    return ans + minimum;
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  }
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}