A proof for the Boyer-Moore Majority Vote algorithm

  • 2

    I wrote the proof for myself. Hope you also find it useful. You might want to read the code first before reading the proof below.

    Since we scan the array again at the end to count the actual appearances of candidate1 and candidate2, what we need to prove is that if a value is one of the majority values, it must equal either candidate1 or candidate2 at the end of the first for loop.

    Let's take a look at the "else" clause of the first for loop. We can consider this step generating a triplet {x, y, z}, where x is the current element of the array that we are processing, y is one of the element with value candidate1 and z is one of the element with value candidate2. Note that we can prove that the values of x, y and z must be different from each other. When the first for loop is completed, it is obvious that every element of the input array is in exactly one of the triplet, except for count1 of the elements with value candidate1 and count2 of the elements with candidate2. These count1 + count2 elements are not in any triplets.

    Let's assume that m is a majority value and m equals neither candidate1 nor candidate2 when the loop is completed. Let's say value m appears k times in the input array. Thus, k > floor(n/3) by definition, where n is the size of the input array. Since the three elements of a triplet must have different values, there must be exactly k triplets generated in the above-mentioned "else" clause each of which contains an element with value m. Obviously, these m triplets contain totally 3k > n different elements, which contradicts the fact that the input array has only n elements. Therefore, m must equal candidate1 or candidate2 when the loop is completed.

    class Solution {
        vector<int> majorityElement(vector<int>& nums) {
            int candidate1 = 0, count1 = 0;
            int candidate2 = 1, count2 = 0;
            // First scan.
            for (const auto &x : nums) {
                if (candidate1 == x)
                else if (candidate2 == x)
                else if (count1 == 0)
                    candidate1 = x;
                    count1 = 1;
                else if (count2 == 0)
                    candidate2 = x;
                    count2 = 1;
                    // Generate a "triplet".
            vector<int> rv;
            count1 = 0;
            count2 = 0;
            // Second scan.
            for (const auto &x : nums) {
                if (x == candidate1)
                else if (x == candidate2)
            if (count1 > int(nums.size() / 3))
            if (count2 > int(nums.size() / 3))
            return rv;

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