Tuesday, July 25th

 Finding the peak in a mountain array is today's problem

An array is designed like a mountain, where values increase from beginning to a peak, when values decline to the end of the array.  We are tasked with finding the peak of the mountain.  The solution must be logarithmic time, because finding the max in an array is a trivial linear time one-liner

Locating a value in logarithmic time beckons binary search

Instead of using binary search to look for a target value, we are looking for a peak.  The peak, which is guaranteed to exist, occurs when a value has both a right and left neighbor with a smaller value.  So we modify binary search to test for the peak condition

Initialize variables to represent the indices of the middle value, left of middle, and right of middle

    int m; // middle
    int mr; // middle right
    int ml; // middle left

Set the index variables to the middle value, left and right.   l and r variables are left and right indices from vanilla binary search

    m=(l+r)/2;
    mr=m+1;
    ml=m-1;

With the middle, middle right, and middle left indices, we test for a peak.  We find the array values at m, ml, and mr, and test if the middle value is greater than the values to the immediate left and right.  If the conditions are met, were return m, the index of the peak;

if( arr[m]>arr[ml] && arr[m]>arr[mr] ) return m; 

The full algorithm is below:

int peakIndexInMountainArray(vector<int>& arr) {
        int l=0;
        int r=arr.size();
        int m; // middle
        int mr; // middle right
        int ml; // middle left
        while (l<r) {
            m=(l+r)/2;
            mr=m+1;
            ml=m-1;
            if( arr[m]>arr[ml] && arr[m]>arr[mr] ) return m;
            else if ( arr[ml]<arr[m]) {
                l=m;
            } else {
                r=m;
            }
        }
        return -1;
    }

The algorithm returns -1 at the end because it was required by the compiler, but since a peak is guaranteed to exist, and the array is of at least length 3 so we don't need to test boundary conditions, the return -1 should never occur, which held up for all test cases

The problem is listed as a medium, but can be solved with binary search by basically modifying one line

Comments

Post a Comment

Popular posts from this blog

Wednesday, September 20th

Image
 Reduce X to Zero Given an integer array and value x, use elements at the right and left of the array to reduce the x value to zero A collection of windows, by Daderot .  Shared under the Creative Commons Attribution-Share Alike 3.0 Unported License .  No changes made A key observation is that if elements at the ends of the array are used to sum up to a target value x , then elements in a contiguous subarray will equal the total_sum - x .  So we will look for the largest window equal to  total_sum - x Since all elements in the array are positive, then we can solve the problem using 2 pointers that form a sliding window. The general idea is to begin 2 pointers right and left at index 0.  Continue incrementing the right index by 1, adding up the elements it passes into the current_window  sum.  If the sum equals the target value, update the length.  Otherwise, if the sum is greater than the target sum, increment the left index until the window...
Read more

Monday, September 11th

Image
 Groups There are a given number of people, each of whom is assigned to a group.  The number identifier of the group is also its size.  The size of a group may be a multiple of its size, e.g. a group of 3 may have 6 people.  Return the  collection of all groups . Team USA Softball (a group).  Shared by USASoftball , via Wikipedia , distributed under the Creative Commons Attribution-Share Alike 3.0 Unported license.  No changes made. This problem is pretty straightforward.  The idea is to create a map (or a vector) keyed by the group, storing an array of individuals belonging to the group. First we create the vector-of-vectors to store our solution, then we initialize a vector of vectors that will store individuals according to their group numbers.  The size of the solution vector is n+1, because contrary to what the beginning of the problem statement said, the indices/group sizes go from 1 to n, according to the stated constraints. vect...
Read more

Tuesday, September 19th

Image
 Find the duplicate number Given an array of integers of length n+1, which contains numbers within the range [1,n], and contains on element that is duplicated at least once, find the duplicate element , considering use of only constant space and no changes to the array "How They Met Themselves" by Dante Rossetti, depicting doppelgängers The problem of finding duplicates is made difficult by the constraints of no modifications to the original array, and use of only constant extra space.  Still, despite these constraints there are still several ways to solve the problem.   Since yesterday's solution involved binary search, we will hone our BS skills by applying it to today's problem The general idea is based on the observation that without duplicates, the number of elements less than or equal an element is equal to the element itself.  E.g., in the collection {1,2,3,4}, the number of elements less than or equal to 2 is 2, less than or equal to 3 is 3, etc.   Howe...
Read more