HackerRank Almost Equal – Advanced problem solution YASH PAL, 31 July 2024 In this HackerRank Almost Equal – Advanced problem solution, we have given an array H[], where H[i] represents the height of the ith fighter, for a given l, r where 0 <= l <= r < N, can you count the number of pairs of fighters between l and r (both inclusive) who qualify to play a game?. Problem solution in Java Programming. import java.io.*; import java.util.*; public class Solution { static int nn; static int block; static int[] fenwick; static long cnt = 0; static class B implements Comparable<B> { int l; int r; int id; @Override public int compareTo(B o) { int i = l/block; int j = o.l/block; if (i != j) { return i - j; } if (r != o.r) { return r - o.r; } return l - o.l; } } static int getSum(int x) { int s = 0; for (; x != 0; x &= x-1) s += fenwick[x-1]; return s; } static class Node { int l; int r; int h; public Node(int l, int r, int h) { this.l = l; this.r = r; this.h = h; } void remove() { add(h, -1); cnt -= getSum(r) - getSum(l); } void add() { cnt += getSum(r) - getSum(l); add(h, 1); } void add(int x, int v) { for (; x < nn; x |= x+1) fenwick[x] += v; } } static public int lowerBound(int[] arr, int len, int key) { if (key <= arr[0]) { return 0; } if (key > arr[len - 1]) { return 0; } int index = Arrays.binarySearch(arr, 0, len, key); if (index < 0) { index = - index - 1; if (index < 0) { return 0; } } return index; } static public int upperBound(int[] arr, int len, int key) { int index = Arrays.binarySearch(arr, 0, len, key+1); if (index < 0) { index = - index - 1; if (index < 0 || index > len) { return 0; } } return index; } public static int unique(int[] arr) { if (arr.length == 1) return 1; int len = 1; while (len < arr.length && arr[len] != arr[len-1]) { len++; } for (int i = len + 1; i < arr.length; i++) { if (arr[i] != arr[len - 1]) { len++; arr[len - 1] = arr[i]; } } return len; } public static void main(String[] args) throws IOException { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); BufferedWriter bw = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH"))); StringTokenizer st = new StringTokenizer(br.readLine()); int n = Integer.parseInt(st.nextToken()); int k = Integer.parseInt(st.nextToken()); st = new StringTokenizer(br.readLine()); int[] a = new int[n]; int[] c = new int[n]; for (int i = 0; i < n; i++) { int item = Integer.parseInt(st.nextToken()); c[i] = a[i] = item; } Arrays.sort(c); nn = unique(c); Node[] nodes = new Node[n]; for (int i = 0; i < n; i++) { int l = lowerBound(c, nn, a[i]-k); int r = upperBound(c, nn, a[i]+k); int h = lowerBound(c, nn, a[i]); nodes[i] = new Node(l, r, h); } st = new StringTokenizer(br.readLine()); int q = Integer.parseInt(st.nextToken()); B[] b = new B[q]; block = (int) Math.max(1.0, Math.sqrt((double)(n)*n/Math.max(1, q))); for (int i = 0; i < q; i++) { st = new StringTokenizer(br.readLine()); b[i] = new B(); b[i].id = i; b[i].l = Integer.parseInt(st.nextToken()); b[i].r = Integer.parseInt(st.nextToken()) + 1; } Arrays.sort(b); int l = 0; int r = 0; fenwick = new int[n]; long[] result = new long[q]; for (int i = 0; i < q; i++) { while (l < b[i].l) { nodes[l++].remove(); } while (b[i].l < l) { nodes[--l].add(); } while (b[i].r < r) { nodes[--r].remove(); } while (r < b[i].r) { nodes[r++].add(); } result[b[i].id] = cnt; } for (int i = 0; i < result.length; i++) { bw.write(String.valueOf(result[i])); if (i != result.length - 1) { bw.write("n"); } } bw.newLine(); bw.close(); br.close(); } } Problem solution in C++ programming. #include <cstdio> #include <cmath> #include <algorithm> #include <utility> #define rep(i, s, n) for(int i=int(s); i<=int(n); ++i) #define rf freopen("in.in", "r", stdin) #define wf freopen("out.out", "w", stdout) using namespace std; #define mp make_pair #define ll long long const int mx=1e5+10; #define gc getchar_unlocked void scan(int &x) { register int c = gc(); x = 0; for(;(c<48 || c>57);c = gc()); for(;c>47 && c<58;c = gc()) {x = (x<<1) + (x<<3) + c - 48;} } static ll b2d[410][410], b1d[mx][410]; static int h[mx], szb[410], lb[410], rb[410]; pair<int, int> bb[410], *buck[410]; int n, k, q, sqr; inline void cum2d() { rep(i, 1, 409) rep(j, 1, 409) b2d[i][j] += b2d[i-1][j]+b2d[i][j-1]-b2d[i-1][j-1]; } inline void cum1d() { rep(i, 0, n) rep(j, 1, 409) b1d[i][j] += b1d[i][j-1]; } inline ll inonev(int* v, int sz) { if(sz<=0) return 0; ll ret = 0; int up=0; rep(i, 0, sz-1) { while(v[up] <= v[i]+k and up<sz) up++; ret += up-(i+1); } return ret; } inline ll intwov(int* v1, int sz1, int* v2, int sz2) { if(sz1<=0 or sz2<=0) return 0; ll ret=0; int low=0, up=0; rep(i, 0, sz1-1) { while(v2[low] < v1[i]-k and low<sz2) low++; while(v2[up] <= v1[i]+k and up<sz2) up++; ret += up-low; } return ret; } void process() { for(int i=0; i*sqr<n; ++i) { int s=i*sqr, e=min(n-1, s+sqr-1); buck[i+1]=new pair<int, int>[e-s+1]; szb[i+1]=e-s+1; rep(j, s, e) buck[i+1][j-s] = mp(h[j], j), lb[j-s] = h[j]; sort(lb, lb+szb[i+1]); sort(buck[i+1], buck[i+1]+szb[i+1]); ll val=inonev(lb, szb[i+1]); b2d[i+1][i+1] += val; rep(j, 1, i) { val=0; int up=0, low=0; rep(l, 0, szb[i+1]-1) { while(buck[j][low].first < buck[i+1][l].first-k and low<szb[j]) low++; while(buck[j][up].first <= buck[i+1][l].first+k and up<szb[j]) up++; b1d[buck[i+1][l].second][j] += up-low; val += up-low; } b2d[j][i+1]+=val; } } for(int i=0; i*sqr<=n; ++i) { for(int j=i+2; szb[j]; ++j) { int up=0, low=0; rep(l, 0, szb[i+1]-1) { while(buck[j][low].first < buck[i+1][l].first-k and low<szb[j]) low++; while(buck[j][up].first <= buck[i+1][l].first+k and up<szb[j]) up++; b1d[buck[i+1][l].second][j] += up-low; } } } } int main() { //rf; wf; scan(n); scan(k); sqr=sqrt(n); rep(i, 0, n-1) scan(h[i]); process(); cum1d(); cum2d(); scanf("%d", &q); while(q--) { int l, r; scan(l); scan(r); ll ans=0; int sqrl=l/sqr, sqrr=r/sqr, szl=0, szr=0; sqrl++; rep(i, 0, szb[sqrl]-1) if(buck[sqrl][i].second>=l and buck[sqrl][i].second<=r) lb[szl++]=buck[sqrl][i].first; rep(i, 0, szb[sqrr+1]-1) if(buck[sqrr+1][i].second>=l and buck[sqrr+1][i].second<=r) rb[szr++]=buck[sqrr+1][i].first; ans+=inonev(lb, szl); if(sqrr+1==sqrl) {printf("%lldn", ans); continue;} ans+=inonev(rb, szr); ans += intwov(lb, szl, rb, szr); ans += b2d[sqrr][sqrr]-b2d[sqrr][sqrl]-b2d[sqrl][sqrr]+b2d[sqrl][sqrl]; rep(i, l, sqr*sqrl-1) ans += b1d[i][sqrr]-b1d[i][sqrl]; rep(i, sqr*sqrr, r) ans += b1d[i][sqrr]-b1d[i][sqrl]; printf("%lldn", ans); } return 0; } Problem solution in C programming. #include <stdio.h> #include <stdlib.h> #include <math.h> void QQ(int x,int y); void add(int X); void removee(int X); void sort_a2(int*a,int*b,int size); void merge2(int*a,int*left_a,int*right_a,int*b,int*left_b,int*right_b,int left_size,int right_size); void sort_a3(int*a,int*b,int*c,int size); void merge3(int*a,int*left_a,int*right_a,int*b,int*left_b,int*right_b,int*c,int*left_c,int*right_c,int left_size,int right_size); int read(int idx); void update(int idx,int val,int MaxVal); int H[300000],rH[300000],q[2][100000],qq[2][100000],idx[300000],tree[300001],NN,cl,cr; long long ans[100000],x; int main(){ int K,Q,i,j; scanf("%d%d",&NN,&K); for(i=0;i<NN;i++){ scanf("%d",&H[3*i]); H[3*i+1]=H[3*i]-K; H[3*i+2]=H[3*i]+K; idx[3*i]=3*i; idx[3*i+1]=3*i+1; idx[3*i+2]=3*i+2; } sort_a2(H,idx,3*NN); for(i=j=0;i<3*NN;i++){ if(i && H[i]!=H[i-1]) j++; rH[idx[i]]=j; } scanf("%d",&Q); for(i=0;i<Q;i++){ scanf("%d%d",&q[0][i],&q[1][i]); q[1][i]++; } for(i=0,j=(int)sqrt(NN);i<Q;i++){ qq[0][i]=q[0][i]/j; qq[1][i]=q[1][i]; idx[i]=i; } sort_a3(&qq[0][0],&qq[1][0],idx,Q); for(i=cl=cr=0,x=0;i<Q;i++){ QQ(q[0][idx[i]],q[1][idx[i]]); ans[idx[i]]=x; } for(i=0;i<Q;i++) printf("%lldn",ans[i]); return 0; } void QQ(int x,int y){ while(cl<x) removee(cl++); while(cl>x) add(--cl); while(cr<y) add(cr++); while(cr>y) removee(--cr); return; } void add(int X){ int y=read(rH[3*X+2]+1); if(rH[3*X+1]) y-=read(rH[3*X+1]); x+=y; update(rH[3*X]+1,1,3*NN); return; } void removee(int X){ int y=read(rH[3*X+2]+1); if(rH[3*X+1]) y-=read(rH[3*X+1]); x-=y-1; update(rH[3*X]+1,-1,3*NN); return; } void sort_a2(int*a,int*b,int size){ if (size < 2) return; int m = (size+1)/2,i; int*left_a,*left_b,*right_a,*right_b; left_a=(int*)malloc(m*sizeof(int)); right_a=(int*)malloc((size-m)*sizeof(int)); left_b=(int*)malloc(m*sizeof(int)); right_b=(int*)malloc((size-m)*sizeof(int)); for(i=0;i<m;i++){ left_a[i]=a[i]; left_b[i]=b[i]; } for(i=0;i<size-m;i++){ right_a[i]=a[i+m]; right_b[i]=b[i+m]; } sort_a2(left_a,left_b,m); sort_a2(right_a,right_b,size-m); merge2(a,left_a,right_a,b,left_b,right_b,m,size-m); free(left_a); free(right_a); free(left_b); free(right_b); return; } void merge2(int*a,int*left_a,int*right_a,int*b,int*left_b,int*right_b,int left_size,int right_size){ int i = 0, j = 0; while (i < left_size|| j < right_size) { if (i == left_size) { a[i+j] = right_a[j]; b[i+j] = right_b[j]; j++; } else if (j == right_size) { a[i+j] = left_a[i]; b[i+j] = left_b[i]; i++; } else if (left_a[i] <= right_a[j]) { a[i+j] = left_a[i]; b[i+j] = left_b[i]; i++; } else { a[i+j] = right_a[j]; b[i+j] = right_b[j]; j++; } } return; } void sort_a3(int*a,int*b,int*c,int size){ if (size < 2) return; int m = (size+1)/2,i; int *left_a,*left_b,*left_c,*right_a,*right_b,*right_c; left_a=(int*)malloc(m*sizeof(int)); right_a=(int*)malloc((size-m)*sizeof(int)); left_b=(int*)malloc(m*sizeof(int)); right_b=(int*)malloc((size-m)*sizeof(int)); left_c=(int*)malloc(m*sizeof(int)); right_c=(int*)malloc((size-m)*sizeof(int)); for(i=0;i<m;i++){ left_a[i]=a[i]; left_b[i]=b[i]; left_c[i]=c[i]; } for(i=0;i<size-m;i++){ right_a[i]=a[i+m]; right_b[i]=b[i+m]; right_c[i]=c[i+m]; } sort_a3(left_a,left_b,left_c,m); sort_a3(right_a,right_b,right_c,size-m); merge3(a,left_a,right_a,b,left_b,right_b,c,left_c,right_c,m,size-m); free(left_a); free(right_a); free(left_b); free(right_b); free(left_c); free(right_c); return; } void merge3(int*a,int*left_a,int*right_a,int*b,int*left_b,int*right_b,int*c,int*left_c,int*right_c,int left_size,int right_size){ int i = 0, j = 0; while (i < left_size|| j < right_size) { if (i == left_size) { a[i+j] = right_a[j]; b[i+j] = right_b[j]; c[i+j] = right_c[j]; j++; } else if (j == right_size) { a[i+j] = left_a[i]; b[i+j] = left_b[i]; c[i+j] = left_c[i]; i++; } else if (left_a[i] == right_a[j]) { if(left_b[i]<=right_b[j]){ a[i+j] = left_a[i]; b[i+j] = left_b[i]; c[i+j] = left_c[i]; i++; } else{ a[i+j] = right_a[j]; b[i+j] = right_b[j]; c[i+j] = right_c[j]; j++; } } else if (left_a[i] < right_a[j]) { a[i+j] = left_a[i]; b[i+j] = left_b[i]; c[i+j] = left_c[i]; i++; } else { a[i+j] = right_a[j]; b[i+j] = right_b[j]; c[i+j] = right_c[j]; j++; } } return; } int read(int idx){ int sum = 0; while(idx>0){ sum+=tree[idx]; idx-=(idx&-idx); } return sum; } void update(int idx,int val,int MaxVal){ while(idx<=MaxVal){ tree[idx]+=val; idx+=(idx&-idx); } return; } coding problems data structure