Separate Even and Odd Numbers
Medium
C++
Arrays
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Problem Statement
Rearrange array so all even numbers come before odd numbers. This teaches array partitioning logic.
Logic: Create two arrays for even/odd, then merge back
Logic: Create two arrays for even/odd, then merge back
Input Format
An integer n (size of array)
Then n integers.
Then n integers.
Output Format
Array rearranged so that:
• All even numbers come first
• All odd numbers come after
(Order inside even and odd groups is maintained)
• All even numbers come first
• All odd numbers come after
(Order inside even and odd groups is maintained)
Example
7 1 2 3 4 5 6 7
2 4 6 1 3 5 7
Constraints
• n ≥ 1
• Use extra arrays for even and odd
• Use extra arrays for even and odd
Concept Explanation
Even and odd numbers are separated first, then combined back into the original array.
Step-by-Step Explanation
1.Create an empty array even[].
2.Create an empty array odd[].
3.Traverse the original array from start to end.
4.For each element:
• If element is even, add it to even[].
• Else, add it to odd[].
5.Copy all elements of even[] back to the original array.
6.After that, copy all elements of odd[] to the array.
7.Print the final array.
2.Create an empty array odd[].
3.Traverse the original array from start to end.
4.For each element:
• If element is even, add it to even[].
• Else, add it to odd[].
5.Copy all elements of even[] back to the original array.
6.After that, copy all elements of odd[] to the array.
7.Print the final array.
Concept Explanation
Even and odd numbers are separated first, then combined back into the original array.
Step-by-Step Explanation
1.Create an empty array even[].
2.Create an empty array odd[].
3.Traverse the original array from start to end.
4.For each element:
• If element is even, add it to even[].
• Else, add it to odd[].
5.Copy all elements of even[] back to the original array.
6.After that, copy all elements of odd[] to the array.
7.Print the final array.
2.Create an empty array odd[].
3.Traverse the original array from start to end.
4.For each element:
• If element is even, add it to even[].
• Else, add it to odd[].
5.Copy all elements of even[] back to the original array.
6.After that, copy all elements of odd[] to the array.
7.Print the final array.
Input / Output Format
Input Format
An integer n (size of array)
Then n integers.
Then n integers.
Output Format
Array rearranged so that:
• All even numbers come first
• All odd numbers come after
(Order inside even and odd groups is maintained)
• All even numbers come first
• All odd numbers come after
(Order inside even and odd groups is maintained)
Constraints
• n ≥ 1
• Use extra arrays for even and odd
• Use extra arrays for even and odd
Examples
Input:
7
1 2 3 4 5 6 7
Output:
2 4 6 1 3 5 7
Example Solution (Public)
void question9_separate_even_odd() {
int arr[] = {12, 7, 18, 5, 20, 9};
int size = 6;
int result[6];
int index = 0;
// First add all even numbers
for(int i = 0; i < size; i++) {
if(arr[i] % 2 == 0) {
result[index++] = arr[i];
}
}
// Then add all odd numbers
for(int i = 0; i < size; i++) {
if(arr[i] % 2 != 0) {
result[index++] = arr[i];
}
}
cout << "Separated array: ";
for(int i = 0; i < size; i++) {
cout << result[i] << " ";
}
cout << endl;
}
Official Solution Code
void question9_separate_even_odd() {
int arr[] = {12, 7, 18, 5, 20, 9};
int size = 6;
int result[6];
int index = 0;
// First add all even numbers
for(int i = 0; i < size; i++) {
if(arr[i] % 2 == 0) {
result[index++] = arr[i];
}
}
// Then add all odd numbers
for(int i = 0; i < size; i++) {
if(arr[i] % 2 != 0) {
result[index++] = arr[i];
}
}
cout << "Separated array: ";
for(int i = 0; i < size; i++) {
cout << result[i] << " ";
}
cout << endl;
}
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