---

Finding the Majority Element

---

• The Majority Element Problem
  • Problem Description:

    Assume that an integer array A[1..n] has a majority element Elements other than majority element need not be distinct. Problem: find majority element.

  • Example:

    A[1..N] = 1 4 5 6 4 4 4 7 2 3 4 4 4

    The majority element in the example is 4.

  ---

  ---

• The naive solution
  • Naive solution:

    Q = ∅ // Q is a list of the form <Value, Count> for ( i = 1; i <= N; i++ ) if ( A[i] ∈ Q ) { increase A[i]'s count by 1; } else { insert <A[i], 1> in Q; } for ( each e ∈ Q ) if ( e.count > N/2 ) print e;

  ---

  ---

• An interesting algorithm
  • Consider the following algorithm:

    winner = A[1]; // assume A[1] is the winner... votes = 1; // A[1] has 1 "surplus" vote so far for ( i = 2; i <= N; i++ ) { if ( A[i] == winner ) { votes++; // One more vote for the assumed winner } else if ( votes > 0 ) { votes--; // One vote against the assumed winner... } else { // The assumed winner has no more surplus votes... winner = A[i]; // Pick a new assumed winner votes = 1; // A[i] has 1 "surplus" vote so far } } return winner

    ---

  • Example Program: (Demo above code)                                                
    • Prog file: click here

    ---

  • Why does the program work:

    Look at the problem as voting for a number of candidates If there is a majority winner, then all the other candidates together does not have the majority vote. I.e.: counting all other candidates votes merged together, it will still fail to defeat the majority element candidate

  ---

  ---

• Why do we study this strange algorithm ?
  • The find frequent elements is a data stream developed by Karp, Papadimitriou and Shenkar is based on this "funny" counting method...