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This approach involves scanning the array with a single pass. Given that an element occurring more than 25% will appear as a block, we can count the occurrences of an element until it changes and check the count.
For a given element arr[i], count consecutive occurrences. If it exceeds the threshold (n/4), return that element immediately.
Time Complexity: O(n), as it makes a single pass through the array. Space Complexity: O(1), uses a constant amount of space.
1#include <stdio.h>
2
3int findSpecialInteger(int* arr, int arrSize) {
4 int threshold = arrSize / 4;
5 int count = 1;
6 for (int i = 1; i < arrSize; i++) {
7 if (arr[i] == arr[i-1]) {
8 count++;
9 if (count > threshold) return arr[i];
10 } else {
11 count = 1;
12 }
13 }
14 return arr[0]; // Fallback case
15}
16
17int main() {
18 int arr[] = {1,2,2,6,6,6,6,7,10};
19 int result = findSpecialInteger(arr, 9);
20 printf("%d", result);
21 return 0;
22}This C program uses a for loop to iterate over the sorted array. Using a threshold of n/4, it checks for consecutive elements, increasing a count until the count exceeds the threshold, returning the element if found.
This approach involves using binary search to take advantage of the sorted nature of the array. By checking elements at regular intervals (n/4), we can identify any elements that repeat in significant blocks more than the threshold.
Instead of scanning the entire array sequentially, we check index positions at specific intervals (n/4, n/2, 3n/4) to find the repeating element.
Time Complexity: O(1) average on small fixed comparisons. Space Complexity: O(1).
1
This JavaScript function undertakes a strategic interval comparison campaign at 25% sections to identify if any element consistently appears across these checkpoints more than the required quota, securing a swift determination.