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Complete guide to searching algorithms — Linear Search, Binary Search, variations, two-pointer technique, and search problems with Java implementations.
Linear Search — O(n)
Example
public static int linearSearch(int[] arr, int target) {
for (int i = 0; i < arr.length; i++) {
if (arr[i] == target) return i;
}
return -1; // Not found
}
Binary Search — O(log n)
Prerequisite: Array must be sorted.
| Step 1: left=0, right=6, mid=3 | arr[3]=7 → FOUND! |
| Step 1: left=0, right=6, mid=3 | arr[3]=7 < 9 → search right |
| Step 2: left=4, right=6, mid=5 | arr[5]=11 > 9 → search left |
| Step 3: left=4, right=4, mid=4 | arr[4]=9 → FOUND! |
Example
// Iterative Binary Search
public static int binarySearch(int[] arr, int target) {
int left = 0, right = arr.length - 1;
while (left <= right) {
int mid = left + (right - left) / 2; // Avoid overflow
if (arr[mid] == target) return mid;
else if (arr[mid] < target) left = mid + 1;
else right = mid - 1;
}
return -1;
}
// Recursive Binary Search
public static int binarySearchRecursive(int[] arr, int target, int left, int right) {
if (left > right) return -1;
int mid = left + (right - left) / 2;
if (arr[mid] == target) return mid;
if (arr[mid] < target) return binarySearchRecursive(arr, target, mid + 1, right);
return binarySearchRecursive(arr, target, left, mid - 1);
}
Two Pointer Technique
Example
// Two Sum (sorted array)
public static int[] twoSum(int[] arr, int target) {
int left = 0, right = arr.length - 1;
while (left < right) {
int sum = arr[left] + arr[right];
if (sum == target) return new int[]{left, right};
else if (sum < target) left++;
else right--;
}
return new int[]{-1, -1};
}
Interview Questions
Q1: When would you use Linear Search over Binary Search?
Answer: When data is unsorted (sorting for binary search costs O(n log n) — not worth it for single search), small arrays (n < 20), or linked lists (no random access for binary search).
Q2: Why use left + (right - left) / 2 instead of (left + right) / 2?
Answer: To prevent integer overflow. If left and right are both near Integer.MAX_VALUE, their sum overflows to a negative number. The subtraction form avoids this.
Q3: What is the time complexity of Binary Search?
Answer: O(log n). Each comparison eliminates half the remaining elements. For n=1,000,000 it takes at most 20 comparisons (log₂(1,000,000) ≈ 20).
Q4: How do you find the peak element in a mountain array?
Answer: Binary search: if arr[mid] < arr[mid+1], peak is to the right (left = mid+1). Otherwise, peak is at mid or left (right = mid). Time: O(log n).
Q5: Can Binary Search be applied to non-array problems?
Answer: Yes! "Binary search on the answer" applies when: the answer space is monotonic (sorted), and you can verify a candidate answer in O(n) or less. Examples: minimum capacity, maximum minimum distance, finding sqrt.
Exam Focus
Revise definitions, diagrams, examples, and short-answer points for Searching Algorithms in Java.
Interview Use
Prepare one clear explanation, one practical example, and one common mistake for this Java Master Course topic.
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