Given an integer array sorted in non-decreasing order, there is exactly one integer in the array that occurs more than 25% of the time, return that integer.
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| Example 1:
Input: arr = [1,2,2,6,6,6,6,7,10]
Output: 6
Example 2:
Input: arr = [1,1]
Output: 1
|
Constraints:
- 1 <= arr.length <= 10^4
- 0 <= arr[i] <= 10^5
Solution#
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| class Solution {
public int findSpecialInteger2(int[] arr) {
int n = arr.length;
int val1 = arr[n / 4];
int val2 = arr[n / 4 * 2] ;
int val3 = arr[n / 4 * 3];
int val4 = arr[n - 1];
int count1 = countFreq(arr, val1);
int count2 = countFreq(arr, val2);
int count3 = countFreq(arr, val3);
int count4 = countFreq(arr, val4);
if (count1 >= count2 && count1 >= count3 && count1 >= count4) {
return val1;
}
if (count2 >= count1 && count2 >= count3 && count2 >= count4) {
return val2;
}
if (count3 >= count1 && count3 >= count2 && count3 >= count4) {
return val3;
}
return val4;
}
int countFreq(int[] arr, int val) {
int mid = Arrays.binarySearch(arr, val);
int right = mid;
while (right < arr.length - 1 && arr[right] == arr[right + 1]) {
right++;
}
int left = mid;
while (left > 0 && arr[left] == arr[left - 1]) {
left--;
}
return right - left;
}
}
|
Solution 2#
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| class Solution {
public int findSpecialInteger(int[] arr) {
int n = arr.length;
int t = n / 4;
for (int i = 0; i < n - t; i++) {
if (arr[i] == arr[i + t]) {
return arr[i];
}
}
return -1;
}
}
|