Chapter 3, Problem 3C

Explanation of Solution

Algorithm:

traverse the list for i=0 to n-1 elements

{

Check for sign of A[abs(A[i])];

If positive then

  Make it negative by A[abs(A[i])]=-A[abs(A[i])];

Else //that is A[abs(A[i])] is negative

  this element (ith element of list) is a repetition

}

Explanation:

The above algorithm is used to find the repeated integer in an array of size n>=2...