Arrays & Strings

Find the duplicate elements in an array of size n where each element is in the range 0 to n-1.

Approach 1 :: Compare each element of the array with all other elements (using two loops) [ O(n2) solution ( not an efficient one) ]
Approach 2 :: Maintain a hash table. Set the hash value to 1 if we encounter the element for the first time. If we see it second time ( we check it by looking at the hash value ), print it [ efficient in terms of time O(n) but additional space is required ]
Approach 3 :: We will exploit the constraintevery element is in the range 0 to n-1“.
Suppose we see an element 4 in the array. We will make arr[4] negative.
So while traversing the array, if we see a negative value, we will print the index of that value.
Considering our example, we will find arr[4] negative and so we will print 4.

using namespace std;

// find all the duplicate elements in an array of size
// 'n' where all elements are in the range 0 to n-1
// If an element 'k' is present in an array, we will
// make the value at index 'k' negative
// i.e if k = 4 then arr[k] = -arr[k]
void findDuplicates(int arr[],int size) {
   int i;
   int k;
   for (i=0;i<size;i++) {
      k = abs(arr[i]);
      if (arr[k] >= 0) {
         // k was not seen previously
         // mark k as seen
         arr[k] = -arr[k];
      else {
         // arr[k] is negative
         // this implies 'k' was seen previously
         cout<<k<<" ";

// main
int main() {
   int arr[] = {3,5,1,6,5,7,3,6};
   int size = sizeof(arr)/sizeof(arr[0]);
   return 0;

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* Find the smallest positive element missing in the given array of size n consisting of both positive & negative integers
* Given an array of n elements, find max( j – i ) s.t arr(j) > arr(i)
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