patterncMinor
A recursive function that performs a binary search
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searchfunctionperformsrecursivebinarythat
Problem
I created a recursive function that uses binary search to just return true if it finds the value and false if it does not.
I'm new to recursion and binary search so please let me know where I can improve upon.
I'm new to recursion and binary search so please let me know where I can improve upon.
/**
* Uses binary search O(log n). Returns true if the values is in the value array false if it's not. Recursive
*/
bool binarySearch( int value, int values[], int left, int right ) {
//keep track of when the array will be empty and return false
if ( right < left ) {
return false;
}
//Find the middle of the array
int mid = ( left + right ) / 2;
//compare the value to the middle of the array. If it's a match return true, else move the left position and right position accordingly
if( value == values[ mid ] ) {
return true;
} else if ( value < values[ mid ] ) {
right = mid - 1;
} else {
left = mid + 1;
}
//return the function
return binarySearch( value, values, left, right );
}Solution
You return a boolean to indicate whether the value was found or not. The function would be more useful if you returned an index at which the value was found (or -1 if not found). It's better information for the same amount of work.
It appears that
You've implemented the search using recursion, but it could also be done using just a loop. You would avoid the overhead of function calls and the possibility of stack overflow.
It appears that
left is the leftmost index to consider, and right is the rightmost index to consider. It would be more idiomatic to follow the convention of inclusive-exclusive bounds, with left being the leftmost index and right being just beyond the last element. That way, right - left indicates the number of elements in the array.You've implemented the search using recursion, but it could also be done using just a loop. You would avoid the overhead of function calls and the possibility of stack overflow.
int mid = ( left + right ) / 2 is vulnerable to integer overflow if both indices are large. int mid = left + (right - left) / 2 would not overflow.Context
StackExchange Code Review Q#156492, answer score: 3
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