LeetCode 1445: Apples & Oranges Solution
Master LeetCode problem 1445 (Apples & Oranges), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
1445. Apples & Oranges
Problem Explanation
Explanation
To solve this problem, we need to find the maximum number of apples and oranges that can be collected while maintaining a balance between them. We can achieve this by iterating through the given arrays of apples and oranges, and for each position, we calculate the total weight of apples and oranges on both sides of that position. We then update the count of apples and oranges collected based on the balance condition.
Algorithmic Steps:
- Iterate through the given arrays of apples and oranges.
- At each position, calculate the total weight of apples and oranges on both sides of that position.
- Update the count of apples and oranges collected based on the balance condition.
- Return the maximum count of apples and oranges that can be collected.
Time Complexity
The time complexity of this solution is O(n), where n is the total number of elements in the arrays of apples and oranges.
Space Complexity
The space complexity of this solution is O(1) as we are using a constant amount of extra space.
Solution Code
class Solution {
public int maxApplesOranges(int[] apples, int[] oranges) {
int leftApples = 0, rightApples = 0;
int leftOranges = 0, rightOranges = 0;
int maxApples = 0, maxOranges = 0;
for (int i = 0; i < apples.length; i++) {
rightApples += apples[i];
}
for (int i = 0; i < oranges.length; i++) {
rightOranges += oranges[i];
}
for (int i = 0; i < apples.length; i++) {
leftApples += apples[i];
rightApples -= apples[i];
leftOranges += oranges[i];
rightOranges -= oranges[i];
maxApples = Math.max(maxApples, Math.min(leftApples, rightOranges));
maxOranges = Math.max(maxOranges, Math.min(leftOranges, rightApples));
}
return Math.max(maxApples, maxOranges);
}
}Try It Yourself
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