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Find duplicates in O(n) time and O(1) extra space - Set 1 in C++
Suppose we have a list of numbers from 0 to n-1. A number can be repeated as many as possible number of times. We have to find the repeating numbers without taking any extra space. If the value of n = 7, and list is like [5, 2, 3, 5, 1, 6, 2, 3, 4, 5]. The answer will be 5, 2, 3.
To solve this, we have to follow these steps −
for each element e in the list, do the following steps −
- sign := A[absolute value of e]
- if the sign is positive, then make it negative
- Otherwise it is a repetition.
Example
#include<iostream>
#include<cmath>
using namespace std;
void findDuplicates(int arr[], int size) {
for (int i = 0; i < size; i++) {
if (arr[abs(arr[i])] >= 0)
arr[abs(arr[i])] *= -1;
else
cout << abs(arr[i]) << " ";
}
}
int main() {
int arr[] = {5, 2, 3, 5, 1, 6, 2, 3, 4, 1};
int n = sizeof(arr)/sizeof(arr[0]);
findDuplicates(arr, n);
}Output
5 2 3 1
Updated on: 01-Nov-2019
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