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targetIndices.java
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targetIndices.java
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package striver.binarysearch;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
public class targetIndices {
public List<Integer> targetIndices(int[] nums, int target) {
return binarySearch(nums, target);
}
public static List<Integer> binarySearch(int[] nums, int key){
Arrays.sort(nums);
List<Integer> res = new ArrayList();
int low = 0;
int high = nums.length - 1;
int left = -1;
while(low <= high){
int mid = (low + high) / 2;
if(nums[mid] == key){
left = mid;
high = mid - 1;
}else if(nums[mid] > key)
high = mid - 1;
else if(nums[mid] < key)
low = mid + 1;
}
low = 0;
high = nums.length - 1;
int right = -1;
while(low <= high){
int mid = (low + high) / 2;
if(nums[mid] == key){
right = mid;
low = mid + 1;
}else if(nums[mid] > key)
high = mid - 1;
else if(nums[mid] < key)
low = mid + 1;
}
if(left == -1 && right == -1)
return res;
else
while(left <= right){
res.add(left);
left++;
}
// here we cant apply tht typical binary search if we do so tht will end up giving the solution thth will not work in all the test cases
// so for tht we applied binary search for two times once for the left side tht will search for the
// target element in the left side and again for the right side and this will find the target element in the right side
// and tht function will give us the solution tht can be applied on all the test cases..
return res;
}
public static void main(String[]args){
int nums[] = {1 , 2 , 3 , 2 , 5,7};
int target = 2;
List<Integer> ans = binarySearch(nums, target);
System.out.println("These are the indices " + ans +" of the array where the " + target + " element is found");
}
}