HackerEarth Equal Median problem solution

In this HackerEarth Equal Median problem solution You will be given two arrays A and B of odd length N. In one swap operation, you can select a number from A and another number from B and swap them. You need to find the minimum number of swap operations required so that the median of those two arrays becomes equal. The median of the array is the element at the middle position in ascending order.

HackerEarth Equal Median problem solution.

`#include <bits/stdc++.h>using namespace std;#define pb push_back#define FOR(i,x,y) for(LL i = (x) ; i <= (y) ; ++i)#define CLR(x,y) memset(x,y,sizeof(x))#define ALL(x) (x).begin(),(x).end()#define SZ(x) ((LL)(x).size())#define LL long longint N;vector<int> A, B, L;/// Returns number of elements smaller than m in Lint getSmall(vector<int> &L, int m){    int cnt = (lower_bound(ALL(L), m) - L.begin());    return cnt;}/// Returns number of elements greater than m in Lint getLarge(vector<int> &L, int m){    int cnt = (upper_bound(ALL(L), m) - L.begin());    return SZ(L) - cnt;}/// Returns number of elements equals to m in Lint getEqual(vector<int> &L, int m){    int cnt = (upper_bound(ALL(L), m) - lower_bound(ALL(L), m));    return cnt;}int solve(){    sort(ALL(A));    sort(ALL(B));    sort(ALL(L));    int med1 = L[N-1], med2 = L[N];    if(med1 != med2) return -1;    int smallA = getSmall(A, med1);    int smallB = getSmall(B, med1);    int equalA = getEqual(A, med1);    int equalB = getEqual(B, med1);    int largeA = getLarge(A, med1);    int largeB = getLarge(B, med1);    int smallEqualA = smallA + equalA;    int smallEqualB = smallB + equalB;    int largeEqualA = largeA + equalA;    int largeEqualB = largeB + equalB;    int LMX = N/2 + 1;    int A2B = 0, B2A = 0;    if(equalA == 0){        B2A += 1;        equalA += 1;        equalB -= 1;    }    if(equalB == 0){        A2B += 1;        equalB += 1;        equalA -= 1;    }    smallEqualA = smallA + equalA;    smallEqualB = smallB + equalB;    largeEqualA = largeA + equalA;    largeEqualB = largeB + equalB;    if(smallEqualA<LMX){        int x = LMX - smallEqualA;        B2A += x;        smallEqualA += x;        smallEqualB -= x;    }    if(smallEqualB<LMX){        int x = LMX - smallEqualB;        A2B += x;        smallEqualA -= x;        smallEqualB += x;    }    if(largeEqualA<LMX){        int x = LMX - largeEqualA;        B2A += x;        largeEqualA += x;        largeEqualB -= x;    }    if(largeEqualB<LMX){        int x = LMX - largeEqualB;        A2B += x;        largeEqualA -= x;        largeEqualB += x;    }    return max(A2B, B2A);}int main() {    int t, x;    scanf("%d",&t);    FOR(cs,1,t){        A.clear();        B.clear();        L.clear();        scanf("%d",&N);        FOR(i,1,N){            scanf("%d",&x);            A.pb(x);            L.pb(x);        }        FOR(i,1,N){            scanf("%d",&x);            B.pb(x);            L.pb(x);        }        LL res = solve();        printf("%lld\n",res);    }}`