Wednesday, September 20th

 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 sum is less than the target value.

To begin, we initialize the necessary variables

int total=0; // total sum of the array
int n=nums.size(); // length of the array
for(int i=0; i<n; i++){
total+=nums[i];
}
int right, left=0; // pointers
int maxlength=-1; // length of largest window equal to target
int windowsum=0; // sum of the current window
int target=total-x; // target value

Then increment the right pointer by 1, at each step updating the sum of the current window

for(int i=0; i<n; i++) {
right=i;
windowsum+=nums[right];
            ...
        }

If the window sum is greater than the target value, increment the left pointer by 1 and update the sum

        while(windowsum > target && left<=right) {
    windowsum-=nums[left];
left++;
}

Finally, if the sum of the current window equals the target value, update the maximum length of the windows found thus far

for(int i=0; i<n; i++) {
            ...
            if(windowsum == target) {
maxlength=max(maxlength, right-left+1);
}
        }

If no target value was found, return -1.  Otherwise, if a target was found, return the length n less the size of the window, because we are ultimately finding the size of the ends

int minOperations(vector<int>& nums, int x) {
        ...
        if(maxlength==-1) return -1;
else return n-maxlength;
    }

Originally approached this problem as a dynamic programming problem, but got stuck when trying to organize a solution of -1 when the window is empty

Thanks to the Editorial, and comparison to the Maximum Size Subarray Sum Equals k problem

Comments

Post a Comment

Popular posts from this blog

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