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Appendix A: Algorithm Implementations

  • Page ID
    46710
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    Count inversion pair in array range 1 to n

    Problems:

    Give an arrray[1..n], a pair call inversion if with i, j (1 <= i <j <= n) and (array[i] > array[j]) counting how many inversions pair?

    Solution: Use divide and conquer (Hint: same with merge sort)

    C++ source code:

     /**
        * Brute Force
        * Complexity O(n^2)
        *
        **/
        /**
        *
        * Count Inversion by Merger Algorithms
        * Complexity O(nlog(n))
        *
        **/
    
    
    #include <iostream>
    #include <cstdio>
    #include <cassert>
    using namespace std;
    
    const int __OO__ = 1e9 + 7;
    const int SIZE = 1e6 + 5;
    
    int n, a[SIZE];
    
    int brute_force(int A[], int Len) {
        int inv = 0;
        for (int i = 1; i < Len; i ++)
            for (int j = i + 1; j <= Len; j ++)
                if (A[i] > A[j]) ++ inv;
        return inv;
    }
    
    int MERGE_INVERSIONS(int A[], int p, int q, int r) {
        int n1 = q - p + 1,
            n2 = r - q;
        int L[n1 + 1], R[n2 + 1];
        for (int i = 1; i <= n1; i ++)
            L[i] = A[p + i - 1];
        for (int j = 1; j <= n2; j ++)
            R[j] = A[q + j];
        L[n1 + 1] = __OO__, R[n2 + 1] = __OO__;
        int i = 1, j = 1;
        int mInv = 0; bool counted = false;
        for (int k = p; k <= r; k ++) {
            if (!counted && R[j] < L[i]) {
                mInv += n1 - i + 1;
                counted = true;
            }
            if (L[i] <= R[j]) {
                A[k] = L[i];
                i ++;
            } else {
                A[k] = R[j];
                j ++;
                counted = false;
            }
        }
        return mInv;
    }
    int COUNT_INVERSIONS(int A[], int p, int r) {
        int inv = 0;
        if (p < r) {
            int q = p + (r - p) / 2;
            inv += COUNT_INVERSIONS(A, p, q);
            inv += COUNT_INVERSIONS(A, q + 1, r);
            inv += MERGE_INVERSIONS(A, p, q, r);
        }
        return inv;
    }
    int main() {
        assert(freopen("INVERSIONS.INP", "r", stdin));
        cin >> n;
        for (int i = 1; i <= n; i ++)
            cin >> a[i];
        //cout << brute_force(a, n) << endl;
        cout << COUNT_INVERSIONS(a, 1, n) << endl;
        return 0;
    }
    
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