Reduce given array by replacing subarrays of length at least K consisting of even numbers with their length
Last Updated :
31 Jan, 2022
Given an array arr[] of length N, the task is to replace all subarrays of only even elements by their length if the length is greater than or equal to K.
Examples:
Input: arr[] = {3, 6, 10, 2, 7, 6, 4, 8}, K = 2
Output: 3 3 7 3
Explanation: There are two subarrays with consecutive even numbers {6, 10, 2} and {6, 4, 8}.
So they are replaced by their length 3 and 3 respectively.
Input: arr[] = {4, 8, 3, 6, 8}, K=3
Output: 4 8 3 6 8
Explanation: No subarray exists for which length of consecutive even numbers is greater than or equal to 3.
Approach: Follow the below steps to answer this problem:
- Create two vectors, one to store answer ans and one to store consecutive even numbers temp.
- If (K >= N), then return the original vector arr.
- Now, traverse the array, and for each element:
- If the current element is odd, then check if the length of temp is greater or equal to K.
- If it is, then push its length in ans vector.
- Else push the elements of temp in ans vector.
- Clear the temp vector.
- Push the current element in the ans vector.
- If the current element is even, then push it in temp.
- After the loop ends, vector ans contains the final answer.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
vector<int> processArray(vector<int>& arr,
int& K)
{
int N = arr.size(), i, count = 0;
vector<int> ans;
vector<int> temp;
if (K >= N)
return arr;
for (i = 0; i < N; i++) {
if (arr[i] & 1) {
if (temp.size() >= K)
ans.push_back(temp.size());
else {
for (auto& x : temp) {
ans.push_back(x);
}
}
ans.push_back(arr[i]);
temp.clear();
}
else {
temp.push_back(arr[i]);
}
}
if (temp.size() >= K) {
ans.push_back(temp.size());
}
else {
for (auto& x : temp) {
ans.push_back(x);
}
}
return ans;
}
int main()
{
vector<int> arr
= { 3, 6, 10, 2, 7, 6, 4, 8 };
int K = 2;
vector<int> ans = processArray(arr, K);
for (auto& x : ans)
cout << x << " ";
return 0;
}
|
Java
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
class GFG {
static List<Integer> processArray(List<Integer> arr, int K) {
int N = arr.size();
ArrayList<Integer> ans = new ArrayList<Integer>();
ArrayList<Integer> temp = new ArrayList<Integer>();
;
if (K >= N)
return arr;
for (int i = 0; i < N; i++) {
if ((arr.get(i) & 1) > 0) {
if (temp.size() >= K)
ans.add(temp.size());
else {
for (int x : temp) {
ans.add(x);
}
}
ans.add(arr.get(i));
temp.clear();
}
else {
temp.add(arr.get(i));
}
}
if (temp.size() >= K) {
ans.add(temp.size());
} else {
for (int x : temp) {
ans.add(x);
}
}
return ans;
}
public static void main(String args[]) {
List<Integer> arr = Arrays.asList(3, 6, 10, 2, 7, 6, 4, 8);
int K = 2;
List<Integer> ans = processArray(arr, K);
for (int x : ans)
System.out.print(x + " ");
}
}
|
Python3
def processArray(arr, K):
N, count = len(arr), 0
ans = []
temp = []
if (K >= N):
return arr
for i in range(0, N):
if (arr[i] & 1):
if (len(temp) >= K):
ans.append(len(temp))
else:
for x in temp:
ans.append(x)
ans.append(arr[i])
temp.clear()
else:
temp.append(arr[i])
if (len(temp) >= K):
ans.append(len(temp))
else:
for x in temp:
ans.append(x)
return ans
if __name__ == "__main__":
arr = [3, 6, 10, 2, 7, 6, 4, 8]
K = 2
ans = processArray(arr, K)
for x in ans:
print(x, end=" ")
|
C#
using System;
using System.Collections.Generic;
public class GFG {
static List<int> processArray(List<int> arr, int K) {
int N = arr.Count;
List<int> ans = new List<int>();
List<int> temp = new List<int>();
;
if (K >= N)
return arr;
for (int i = 0; i < N; i++) {
if ((arr[i] & 1) > 0) {
if (temp.Count >= K)
ans.Add(temp.Count);
else {
foreach (int x in temp) {
ans.Add(x);
}
}
ans.Add(arr[i]);
temp.Clear();
}
else {
temp.Add(arr[i]);
}
}
if (temp.Count >= K) {
ans.Add(temp.Count);
} else {
foreach (int x in temp) {
ans.Add(x);
}
}
return ans;
}
public static void Main(String []args) {
List<int> arr = new List<int>(new int[]{3, 6, 10, 2, 7, 6, 4, 8});
int K = 2;
List<int> ans = processArray(arr, K);
foreach (int x in ans)
Console.Write(x + " ");
}
}
|
Javascript
<script>
function processArray(arr, K)
{
let N = arr.length, i, count = 0;
let ans = [];
let temp = [];
if (K >= N)
return arr;
for (i = 0; i < N; i++) {
if (arr[i] & 1) {
if (temp.length >= K)
ans.push(temp.length);
else {
for (let x of temp) {
ans.push(x);
}
}
ans.push(arr[i]);
temp = [];
}
else {
temp.push(arr[i]);
}
}
if (temp.length >= K) {
ans.push(temp.length);
}
else {
for (let x of temp) {
ans.push(x);
}
}
return ans;
}
let arr
= [3, 6, 10, 2, 7, 6, 4, 8];
let K = 2;
let ans = processArray(arr, K);
for (let x of ans)
document.write(x + " ");
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(N)
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