Java: Find the index of a value in a sorted array.
Java Basic: Exercise-124 with Solution
Write a Java program to find the index of a value in a sorted array. If the value does not find return the index where it would be if it were inserted in order.
Example:
[1, 2, 4, 5, 6] 5(target) -> 3(index)
[1, 2, 4, 5, 6] 0(target) -> 0(index)
[1, 2, 4, 5, 6] 7(target) -> 5(index)
Pictorial Presentation:

Sample Solution:
Java Code:
import java.util.*;
public class Main {
public static void main(String[] args) {
// Create an array of integers
int[] nums = {1, 2, 4, 5, 6};
int target = 5;
// target = 0;
// target = 7;
// Call the searchInsert function and print the result
System.out.print(searchInsert(nums, target));
}
public static int searchInsert(int[] nums1, int target) {
// Check if the input array is empty or null
if (nums1 == null || nums1.length == 0) {
return 0;
}
// Initialize variables for binary search
int start = 0;
int end = nums1.length - 1;
int mid = start + (end - start) / 2;
while (start + 1 < end) {
mid = start + (end - start) / 2;
// Compare the middle element with the target
if (nums1[mid] == target) {
return mid;
} else if (nums1[mid] > target) {
end = mid;
} else {
start = mid;
}
}
// Determine the insertion position based on binary search results
if (nums1[start] >= target) {
return start;
} else if (nums1[start] < target && target <= nums1[end]) {
return end;
} else {
return end + 1;
}
}
}
Sample Output:
3
Flowchart:

Java Code Editor:
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Next: Write a Java program to get the preorder traversal of its nodes' values of a given a binary tree
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