01(March) Peak element

01. Peak element

The problem can be found at the following link: Question Link

My Approach

To find the peak element within a sequence, we employ a binary search method, leveraging the characteristic pattern where the sequence ascends and then descends.

• Initially, we establish two pointers, left and right, delineating the array's start and end, respectively.

• Next, we iterate through a loop to determine the middle index.

• If the value at the middle index is lower than its right adjacent element, it indicates the peak is positioned towards the right. Consequently, we shift left to mid + 1.

• Conversely, if the middle value is greater than or equal to its adjacent element to the right, it implies the peak lies towards the left or could be the middle value itself. Hence, we adjust right to mid.

• This iterative process continues until left is less than right, signifying that left denotes the peak element within the sequence.

Time and Auxiliary Space Complexity

• Time Complexity: The time complexity of the binary search approach is O(log n), where n is the number of elements in the array.

• Auxiliary Space Complexity: The space complexity is O(1) as we are using only a constant amount of extra space.

Code (C++)

class Solution
{
    public:
    int peakElement(int arr[], int n)
    {
       int ans=0;
       for(int i=0;i<n;i++){
           if(arr[i]>arr[i+1]){
              ans=i;
              break;
           }
          else if(arr[i]>arr[i-1]){
              ans=i;
          }
       }
       return ans;
    }
};

Contribution and Support

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