Ritu Singh
Summary:
Suppose we have N and M as input parameters and return the minimum possible difference between the maximum and minimum number.
Our task is to complete the function findMinDiff(), which takes array A[ ],
Solution:
Explanation:
The method findMinDiff takes three parameters: a vector a of long long integers, representing a sequence of numbers, n representing the size of the sequence, and m representing the number of elements to consider at a time.
Inside the method:
The sort function from the algorithm library is used to sort the input vector a in ascending order.
Variables start and end are initialized to 0 and m - 1 respectively. These variables represent the start and end indices of the subsequence currently being considered.
A variable min_len is initialized to the maximum possible value of a long long integer (INT_MAX), which is used to keep track of the minimum difference found between elements of the subsequence.
A variable dif is initialized to 0, which will store the difference between the current end and start elements of the subsequence.
A while loop runs as long as the end index is less than n. Within this loop:
The difference between the element at the current end index and the element at the current start index is calculated and stored in dif.
The minimum of the current min_len and dif is computed and stored back into min_len.
The start and end indices are incremented to slide the window to the right by one position.
Once the loop ends, the minimum difference (min_len) found among all the subsequences is returned.
Overall, this code efficiently finds the minimum difference between any two elements when considering subsequences of size m from the input sequence a.
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