Maximize Element Fit Count — Problem Statement & Solution Guide
Problem Description
Given two integer arrays, elements and slots, determine the maximum number of elements that can be successfully placed into slots such that each element fits into its assigned slot.
Examples
Input
[1, 2, 3], [1, 1, 1]
Output
1
Explanation: Step-by-step: with input elements = [1, 2, 3] and slots = [1, 1, 1], we first sort both arrays. Then we initialize two pointers, one for elements and one for slots. We iterate through both arrays and for each element, we check if it can fit into the current slot. If it can, we increment the count of successfully placed elements and move to the next slot. In this case, only one element (1) can fit into the first slot, so the output is 1.
Input
[3, 3, 3], [3, 3, 3]
Output
3
Explanation: Step-by-step: with input elements = [3, 3, 3] and slots = [3, 3, 3], we first sort both arrays. Then we initialize two pointers, one for elements and one for slots. We iterate through both arrays and for each element, we check if it can fit into the current slot. If it can, we increment the count of successfully placed elements and move to the next slot. In this case, all three elements (3) can fit into the three slots, so the output is 3.
Constraints
- 1 <= elements.length, slots.length <= 10^5
- 1 <= elements[i], slots[i] <= 10^9
Optimal Approach & Strategy
An optimal approach is to sort both the elements and slots arrays, and then use a two-pointer technique to match elements with slots. This way, we ensure that the largest elements are matched with the largest slots, maximizing the number of elements that fit. This approach has a time complexity of O(n log n) due to the sorting step.
Brute Force Approach
The brute force approach would involve trying each element in each slot, checking if it fits, and then backtracking to explore other possibilities. This would result in a lot of repeated work and be very inefficient. It would involve generating all permutations of elements and slots.
Verified Code Solutions
function solution(elements, slots) {
elements.sort((a, b) => a - b);
slots.sort((a, b) => a - b);
let count = 0;
let i = 0, j = 0;
while (i < elements.length && j < slots.length) {
if (elements[i] <= slots[j]) {
count++;
i++;
j++;
} else {
j++;
}
}
return count;
}class Solution {
public:
int solution(vector<int>& elements, vector<int>& slots) {
sort(elements.begin(), elements.end());
sort(slots.begin(), slots.end());
int count = 0;
int i = 0, j = 0;
while (i < elements.size() && j < slots.size()) {
if (elements[i] <= slots[j]) {
count++;
i++;
j++;
} else {
j++;
}
}
return count;
}
};class Solution {
public int solution(int[] elements, int[] slots) {
Arrays.sort(elements);
Arrays.sort(slots);
int count = 0;
int i = 0, j = 0;
while (i < elements.length && j < slots.length) {
if (elements[i] <= slots[j]) {
count++;
i++;
j++;
} else {
j++;
}
}
return count;
}
}def solution(elements, slots):
elements.sort()
slots.sort()
count = 0
i, j = 0, 0
while i < len(elements) and j < len(slots):
if elements[i] <= slots[j]:
count += 1
i += 1
j += 1
else:
j += 1
return countfunction solution(elements, slots) {
elements.sort((a, b) => a - b);
slots.sort((a, b) => a - b);
let count = 0;
let i = 0, j = 0;
while (i < elements.length && j < slots.length) {
if (elements[i] <= slots[j]) {
count++;
i++;
j++;
} else {
j++;
}
}
return count;
}Asked in Top Tech Interviews
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