904. Fruit Into Baskets.md

904. Fruit Into Baskets

Difficulty : Medium

Related Topics : Two Pointers

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In a row of trees, the i-th tree produces fruit with type tree[i].

You start at any tree of your choice, then repeatedly perform the following steps:

• 1.Add one piece of fruit from this tree to your baskets. If you cannot, stop.
• 2.Move to the next tree to the right of the current tree. If there is no tree to the right, stop.

Note that you do not have any choice after the initial choice of starting tree: you must perform step 1, then step 2, then back to step 1, then step 2, and so on until you stop.

You have two baskets, and each basket can carry any quantity of fruit, but you want each basket to only carry one type of fruit each.

What is the total amount of fruit you can collect with this procedure?

Example 1:

Input: [1,2,1]
Output: 3
Explanation: We can collect [1,2,1].

Example 2:

Input: [0,1,2,2]
Output: 3
Explanation: We can collect [1,2,2].
If we started at the first tree, we would only collect [0, 1].

Example 3:

Input: [1,2,3,2,2]
Output: 4
Explanation: We can collect [2,3,2,2].
If we started at the first tree, we would only collect [1, 2].

Example 4:

Input: [3,3,3,1,2,1,1,2,3,3,4]
Output: 5
Explanation: We can collect [1,2,1,1,2].
If we started at the first tree or the eighth tree, we would only collect 4 fruits.

Note:

• 1 <= tree.length <= 40000
• 0 <= tree[i] < tree.length

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Solution

same as 159. Longest Substring with At Most Two Distinct Characters

• mine
  • Java
    • Two Pointers Runtime: 9 ms, faster than 78.71%, Memory Usage: 79.5 MB, less than 13.10% of Java online submissions

      // O(N)time
      // O(N)space
      public int totalFruit(int[] tree) {
          int res = 0, l = 0, len = tree.length;
          int[] record = new int[len];
          int distinct = 0;
          for (int i = 0; i < len; i++) {
              if (record[tree[i]] == 0) {
                  distinct++;
              }
              record[tree[i]]++;
              while (distinct > 2) {
                  record[tree[l]]--;
                  if (record[tree[l]] == 0) {
                      distinct--;
                  }
                  l++;
              }
              res = Math.max(res, i - l + 1);
          }
          return res;
      }
      

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• the most votes
  • Runtime: 8 ms, faster than 80.52%, Memory Usage: 77.3 MB, less than 17.18% of Java online submissions

    // O(N)time
    // O(1)space
    public int totalFruit(int[] tree) {
        int res = 0, cur = 0, count_b = 0, a = 0, b = 0;
        for (int c :  tree) {
            cur = c == a || c == b ? cur + 1 : count_b + 1;
            count_b = c == b ? count_b + 1 : 1;
            if (b != c) {a = b; b = c;}
            res = Math.max(res, cur);
        }
        return res;
    }
    

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