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Replace All Elements with Zero in an Array
Problem Statement
Given an array of elements, the task is to replace every element of the array with zero.
This means:
- The size of the array remains the same
- Only the values stored at each index are modified
- After the operation, every index contains 0
Why This Problem Is Important
Although simple, this problem helps in understanding:
- In-place modification of arrays
- Array traversal and index-based updates
- Difference between changing values vs changing structure
- Memory efficiency (no new array required)
This operation is frequently used in:
- Resetting buffers
- Initializing data structures
- Clearing previous results
- Simulation and test-case preparation
Input and Output Format
Input
Array: [5, 10, -3, 8, 12]
Output
Array after replacement: [0, 0, 0, 0, 0]
Key Concept
An array stores values at fixed indices.
By traversing the array and assigning 0 at each index, we overwrite existing values without changing the array size.
Step-by-Step Algorithm
- Traverse the array from index 0 to n-1
- For each index i:
- Assign array[i] = 0
- Print the updated array
Pseudocode
for i = 0 to n-1:
array[i] = 0
Dry Run Example
Original Array: [5, 10, -3, 8, 12]
Index 0 → replace 5 with 0
Index 1 → replace 10 with 0
Index 2 → replace -3 with 0
Index 3 → replace 8 with 0
Index 4 → replace 12 with 0
Final Array: [0, 0, 0, 0, 0]
Time and Space Complexity
| Metric | Value |
|---|---|
| Time Complexity | O(n) |
| Space Complexity | O(1) |
The replacement is done in-place, so no extra memory is used.
Language-wise Implementation
C Implementation
#include
int main() {
int arr[] = {5, 10, -3, 8, 12};
int n = sizeof(arr) / sizeof(arr[0]);
for(int i = 0; i < n; i++) {
arr[i] = 0;
}
printf("Array after replacement: ");
for(int i = 0; i < n; i++) {
printf("%d ", arr[i]);
}
return 0;
}
Output
Array after replacement: 0 0 0 0 0
C++ Implementation
#include
using namespace std;
int main() {
int arr[] = {5, 10, -3, 8, 12};
int n = sizeof(arr) / sizeof(arr[0]);
for(int i = 0; i < n; i++) {
arr[i] = 0;
}
cout << "Array after replacement: ";
for(int i = 0; i < n; i++) {
cout << arr[i] << " ";
}
return 0;
}
Output
Array after replacement: 0 0 0 0 0
Java Implementation
public class Main {
public static void main(String[] args) {
int[] arr = {5, 10, -3, 8, 12};
for(int i = 0; i < arr.length; i++) {
arr[i] = 0;
}
System.out.print("Array after replacement: ");
for(int num : arr) {
System.out.print(num + " ");
}
}
}
Output
Array after replacement: 0 0 0 0 0
Python Implementation
arr = [5, 10, -3, 8, 12]
for i in range(len(arr)):
arr[i] = 0
print("Array after replacement:", arr)
Output
Array after replacement: [0, 0, 0, 0, 0]
C# Implementation
using System;
class Program {
static void Main() {
int[] arr = {5, 10, -3, 8, 12};
for(int i = 0; i < arr.Length; i++) {
arr[i] = 0;
}
Console.Write("Array after replacement: ");
foreach(int num in arr) {
Console.Write(num + " ");
}
}
}
Output
Array after replacement: 0 0 0 0 0
JavaScript Implementation
let arr = [5, 10, -3, 8, 12];
for (let i = 0; i < arr.length; i++) {
arr[i] = 0;
}
console.log("Array after replacement:", arr);
Output
Array after replacement: [0, 0, 0, 0, 0]
Common Mistakes
- Creating a new array unnecessarily
- Forgetting to print the updated array
- Using incorrect loop bounds
- Assuming array size changes after replacement
Interview Variations
- Replace elements with a given value k
- Replace only even or odd elements with zero
- Reset array using built-in functions
- Replace elements conditionally
Detailed Summary
Replacing all elements of an array with zero is a fundamental in-place modification problem. It emphasizes how arrays store data at fixed indices and how those values can be overwritten without altering the array’s size or structure. By performing a single traversal and updating each index directly, the operation remains memory-efficient with constant space usage. This problem forms the basis for understanding array initialization, clearing data structures, and preparing arrays for reuse in real-world programming and algorithmic scenarios.
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