PROBLEM DESCRIPTION
Given a sorted array of integers A (0 based index) of size N, find the starting and the ending position of a given integer B in array A.
Your algorithm’s runtime complexity must be in the order of O(log n).
Return an array of size 2, such that the first element = starting position of B in A and the second element = ending position of B in A, if B is not found in A return [-1, -1].
SOLUTION
We can use binary search two times - first to get the starting index and then to get the end index.
The main condition in binary search which will change is:
For start index:
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if(A[m] == B){
idx = m;
r = m-1;
}
For end index:
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if(A[m] == B){
idx = m;
l = m+1;
}
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public class Solution {
// DO NOT MODIFY THE ARGUMENTS WITH "final" PREFIX. IT IS READ ONLY
public int[] searchRange(final int[] A, int B) {
int startIdx = startIndex(A, B);
if(startIdx == -1) return new int[]{-1, -1};
int endIdx = endIndex(A, B);
return new int[]{startIdx, endIdx};
}
public int endIndex(int[] A, int B){
int n = A.length;
int l = 0;
int r = n-1;
int idx = -1;
while(l<=r){
int m = (l+r)/2;
if(A[m] == B){
idx = m;
l = m+1;
}else if(A[m] > B){
r = m-1;
}else{
l = m+1;
}
}
return idx;
}
public int startIndex(int[] A, int B){
int n = A.length;
int l = 0;
int r = n-1;
int idx = -1;
while(l<=r){
int m = (l+r)/2;
if(A[m] == B){
idx = m;
r = m-1;
}else if(A[m] > B){
r = m-1;
}else{
l = m+1;
}
}
return idx;
}
}