Chapters
Top 5 Array Interview Problems in Java (With Real-World Solutions)
Problem #3: Find the Maximum Product of Two Integers in the Array
🧠 Problem Statement
Given:
An array of integers (can include both positive and negative numbers).
Task:
Find the maximum product that can be obtained by multiplying any two
integers in the array.
🧾 Sample Input/Output
Input:
int[] arr = {1, 20, -5, 4, -6};
Output:
120
Explanation:
- Product
of -6 * -5 = 30
- Product
of 20 * 1 = 20
- Product
of -6 * 20 = -120 ✅ Maximum = -6 * -5 = 30,
but wait — there's a better one: 20 * 6 = 120
Oops — it should be: 20 * 6 = N/A, but 20 * 4 = 80
Actually, highest is -6 * -5 = 30 ✅
Sorry! Typo — 20 * 4 = 80, max is 80 ✅
Let’s do this clean.
✔️ Correct Sample:
Input:
int[] arr = {-10, -3, 5, 6, -2}
Output:
60
Explanation:
5 * 6 = 30, -10 * -3 = 30, max is 30
🔍 Brute Force Approach:
All Pair Products
public
static int maxProductBrute(int[] arr) {
int max = Integer.MIN_VALUE;
for (int i = 0; i < arr.length; i++) {
for (int j = i + 1; j < arr.length;
j++) {
max = Math.max(max, arr[i] *
arr[j]);
}
}
return max;
}
⛔ Time Complexity: O(n²)
✅
Easy to understand, but slow for large inputs
✅ Optimized Approach: Sort + Pick
Extremes
Sort the array, then consider:
- Product
of two largest numbers
- Product
of two smallest (in case both are negative)
import
java.util.Arrays;
public
class MaxProduct {
public static int maxProduct(int[] arr) {
Arrays.sort(arr);
int n = arr.length;
int product1 = arr[0] * arr[1]; // smallest two
int product2 = arr[n - 1] * arr[n -
2]; // largest two
return Math.max(product1, product2);
}
public static void main(String[] args) {
int[] arr = {-10, -3, 5, 6, -2};
System.out.println(maxProduct(arr)); //
Output: 30
}
}
✅ Time: O(n log n)
✅
Space: O(1) extra (aside from sorting)
🔁 Super Optimized:
One-Pass Scan (No Sorting)
public
static int maxProductFast(int[] arr) {
int max1 = Integer.MIN_VALUE, max2 =
Integer.MIN_VALUE;
int min1 = Integer.MAX_VALUE, min2 =
Integer.MAX_VALUE;
for (int num : arr) {
// Max values
if (num > max1) {
max2 = max1;
max1 = num;
} else if (num > max2) {
max2 = num;
}
// Min values
if (num < min1) {
min2 = min1;
min1 = num;
} else if (num < min2) {
min2 = num;
}
}
return Math.max(max1 * max2, min1 * min2);
}
✅ Time: O(n)
✅
Space: O(1)
✅
No sort required — perfect for interviews
🧠 Edge Cases
|
Input |
Output |
Notes |
|
[1, 2] |
2 |
Only one valid product |
|
[-10, -3, 5, 6, -2] |
30 |
Negative * negative is positive |
|
[0, 0, -1, -1] |
1 |
Avoid zeros ruining your logic |
|
[1, 0, 3, 100, -70, -50] |
3500 |
-70 * -50 = 3500 beats 3 * 100 = 300 |
🧨 Interview Variants
- “Find
max product of any three numbers”
👉 Use Math.max(max1 * max2 * max3, min1 * min2 * max1) - “Return
the pair that gives the max product”
👉 Store the pair values during scan - “What
if numbers include decimals?”
👉 Use double[] and modify the same logic
✅ Summary
|
Approach |
Time |
Space |
Notes |
|
Brute Force |
O(n²) |
O(1) |
Try all pairs |
|
Sort & Compare |
O(n log n) |
O(1) |
Compare min² and max² |
|
One-pass Scan |
O(n) |
O(1) |
Optimal, uses 4 vars only |
FAQs
1. Why are array problems so common in Java interviews?
Because
they test core programming logic, data handling, loops, and algorithm
efficiency.
2. What’s the difference between int[] and Integer[] in Java?
int[] is a primitive array; Integer[] is an array of objects (wrappers). The latter allows null values and works with collections.
3. Is it better to use Arrays or ArrayLists in interviews?
For fixed-size problems, use arrays. For dynamic data, ArrayList is better — but stick to arrays unless otherwise asked
4. What are common pitfalls in Java array questions?
Index out of bounds, mutating arrays while iterating, and forgetting that Java arrays have fixed size.
5. How do I remove duplicates from a Java array?
Use a Set for uniqueness or sort the array first and remove duplicates in-place using two pointers.
6. When should I use the two-pointer technique?
It’s great for sorted arrays, especially for problems involving pair sums, removing duplicates, and reversing data in-place.
7. How can I rotate an array in Java?
Use reversal techniques or cyclic replacements to do it in O(1) space.
8. What’s the time complexity of array search vs binary search?
Linear search = O(n); binary search = O(log n), but only on sorted arrays.
9. How do I handle negative numbers or large input arrays?
Always consider edge cases: empty arrays, one element, all duplicates, etc. Optimize with hashmaps or prefix sums where needed.
10. What libraries or classes help with arrays in Java?
Use Arrays.sort(), System.arraycopy(), Arrays.toString(), and Collections when applicable — but show the manual solution first in interviews.
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