Find Peak Element
Problem Statement :
A peak element is an element that is strictly greater than its neighbors. Given a 0-indexed integer array nums, find a peak element, and return its index. If the array contains multiple peaks, return the index to any of the peaks. You may imagine that nums[-1] = nums[n] = -∞. In other words, an element is always considered to be strictly greater than a neighbor that is outside the array. You must write an algorithm that runs in O(log n) time. Example 1: Input: nums = [1,2,3,1] Output: 2 Explanation: 3 is a peak element and your function should return the index number 2. Example 2: Input: nums = [1,2,1,3,5,6,4] Output: 5 Explanation: Your function can return either index number 1 where the peak element is 2, or index number 5 where the peak element is 6. Constraints: 1 <= nums.length <= 1000 -231 <= nums[i] <= 231 - 1 nums[i] != nums[i + 1] for all valid i.
Solution :
Solution in C :
int findPeakElement(int* nums, int numsSize){
int left = 0;
int right = numsSize - 1;
int mid;
if(numsSize == 1)
return 0;
if(numsSize == 2){
if(nums[0] > nums[1])
return 0;
else
return 1;
}
while(left < right){
mid = (left + right)/2;
if(mid == 0)
break;
if(nums[mid] > nums[mid + 1] && nums[mid] > nums[mid - 1])
return mid;
else if(nums[mid] > nums[mid -1])
left = mid + 1;
else
right = mid - 1;
}
if(right < (numsSize - 1) && right > 0){
if(nums[right] > nums[right + 1] && nums[right] > nums[right - 1])
return right;
}
else if(left > 0 && left < (numsSize -1) ){
if(nums[left] > nums[left + 1] && nums[left] > nums[left - 1])
return left;
}
else if(left == 0)
return left;
else
return right;
return 0;
}
Solution in C++ :
class Solution {
public:
int findPeakElement(vector<int>& nums) {
int start=0,end=nums.size()-1;
while(start<end){
int mid=(start+end)/2;
if(mid==0){
return nums[0] >= nums[1] ? 0 : 1;
}
if (mid == end){
return nums[end - 1] >= nums[end - 2] ? end - 1 : end - 2;
}
if(nums[mid]>nums[mid+1] && nums[mid]>nums[mid-1]){
return mid;
}
if(nums[mid]<nums[mid-1]){
end=mid-1;
}
else{
start=mid+1;
}
}
return start;
}
};
Solution in Java :
class Solution {
public int findPeakElement(int[] nums) {
int start = 0, end = nums.length - 1;
while (start < end) {
int mid = (start + end) / 2;
if (mid == 0) {
return nums[0] >= nums[1] ? 0 : 1;
}
if (mid == end) {
return nums[end - 1] >= nums[end - 2] ? end - 1 : end - 2;
}
if (nums[mid] > nums[mid + 1] && nums[mid] > nums[mid - 1]) {
return mid;
}
if (nums[mid] < nums[mid - 1]) {
end = mid - 1;
} else {
start = mid + 1;
}
}
return start;
}
}
Solution in Python :
class Solution(object):
def findPeakElement(self, nums):
start = 0
end = len(nums) - 1
while start < end:
mid = (start + end) // 2
if mid == 0:
return 0 if nums[0] >= nums[1] else 1
if mid == end:
return end - 1 if nums[end - 1] >= nums[end - 2] else end - 2
if nums[mid] > nums[mid + 1] and nums[mid] > nums[mid - 1]:
return mid
if nums[mid] < nums[mid - 1]:
end = mid - 1
else:
start = mid + 1
return start
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