Leetcode 1287: Element Appearing More Than 25% In Sorted Array
Given a sorted integer array, find and return the integer that occurs more than 25% of the time.
Problem
Approach
Steps
Complexity
Input: The input consists of an integer array sorted in non-decreasing order.
Example: Input: arr = [1,2,2,6,6,6,6,7,10]
Constraints:
• 1 <= arr.length <= 104
• 0 <= arr[i] <= 105
Output: The function should return the integer that appears more than 25% of the time in the array.
Example: Output: 6
Constraints:
• It is guaranteed that exactly one integer will appear more than 25% of the time.
Goal: The goal is to find the element that appears more than 25% of the time in the array.
Steps:
• Traverse through the array to count the frequency of each integer.
• Check if any integer appears more than one-fourth of the array length.
• Return the integer that satisfies the condition.
Goal: The constraints ensure that the array is of manageable size and contains valid integer elements.
Steps:
• The length of the array is between 1 and 10^4.
• Each element in the array is between 0 and 10^5.
Assumptions:
• The input array is already sorted in non-decreasing order.
• Input: Input: arr = [1,2,2,6,6,6,6,7,10]
• Explanation: In this case, the integer '6' appears 4 times, which is more than 25% of the length of the array, so the function returns '6'.
Approach: To solve this problem, we need to count the occurrences of each integer in the array and check if any integer appears more than 25% of the time.
Observations:
• Since the array is sorted, repeated elements will be adjacent to each other, making it easier to count occurrences.
• A hashmap or simple iteration through the array could be used to count the occurrences of each integer.
Steps:
• Initialize a map to store the count of each number.
• Iterate through the array and update the frequency count.
• Once the frequency count is updated, check for the integer that exceeds 25% of the array length.
• Return that integer.
Empty Inputs:
• An empty array will never occur since the minimum length of the array is 1.
Large Inputs:
• The solution should efficiently handle arrays with lengths up to 10^4.
Special Values:
• In the case where all elements are the same, that element will automatically be the result.
Constraints:
• Since exactly one element appears more than 25% of the time, there are no corner cases where no element satisfies the condition.
int findSpecialInteger(vector<int>& arr) {
unordered_map<int, int> m;
for(int i = 0; i < arr.size(); i++){
m[arr[i]]++;
}
for(auto i : m){
if(i.second > arr.size() / 4){
return i.first;
}
}
return arr[0];
}
1 : Function Definition
int findSpecialInteger(vector<int>& arr) {
Defines a function to find the special integer that appears more than 25% of the time.
2 : Data Structure
unordered_map<int, int> m;
Declares an unordered map to store the frequency of each integer in the array.
3 : Loop Start
for(int i = 0; i < arr.size(); i++){
Iterates through the array to populate the frequency map.
4 : Increment Frequency
m[arr[i]]++;
Increments the count of the current integer in the frequency map.
5 : Check Frequencies
for(auto i : m){
Iterates through the frequency map to find the special integer.
6 : Condition Check
if(i.second > arr.size() / 4){
Checks if the current integer's frequency exceeds 25% of the array size.
7 : Return Value
return i.first;
Returns the integer if it satisfies the condition.
8 : Default Return
return arr[0];
Returns the first element as a default value if no special integer is found.
Best Case: O(n) - The best case is when the array is short and the element is immediately found.
Average Case: O(n) - In the average case, we iterate through the entire array to count occurrences.
Worst Case: O(n) - The worst case involves iterating through the entire array and counting all frequencies.
Description: The time complexity is O(n) because we have to process each element in the array.
Best Case: O(1) - If only one unique element appears, the space required will be minimal.
Worst Case: O(n) - The worst case occurs when all elements in the array are unique, so we store the frequency of all elements.
Description: The space complexity is O(n) in the worst case because we need to store the frequency of each element in the array.
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