HackerRank Costly Intervals problem solution YASH PAL, 31 July 2024 In this HackerRank Costly Intervals problem, you are given the array and an integer. For each index i from 1 to n, your goal is to find the largest size of any subarray.Problem solution in Java Programming.import java.io.*; import java.util.*; public class Solution { static class Node { int x; long code; public Node(int x, long code) { this.x = x; this.code = code; } } public static int[][] build(int[] a) { int n = a.length; int b = 32 - Integer.numberOfLeadingZeros(n); int[][] ret = new int[b][]; for (int i = 0, l = 1; i < b; i++, l *= 2) { if (i == 0) { ret[i] = a; } else { ret[i] = new int[n - l + 1]; for (int j = 0; j < n - l + 1; j++) { ret[i][j] = Math.min(ret[i - 1][j], ret[i - 1][j + l / 2]); } } } return ret; } public static int rmq(int[][] or, int l, int r) { assert l <= r; if (l >= r) return Integer.MAX_VALUE; int t = 31 - Integer.numberOfLeadingZeros(r - l); return Math.min(or[t][l], or[t][r - (1 << t)]); } 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()); int[] a = new int[n]; int[] ra = new int[n]; st = new StringTokenizer(br.readLine()); for (int i = 0; i < n; i++) { int item = Integer.parseInt(st.nextToken()); a[i] = item; ra[i] = -a[i]; } int[][] stmin = build(a); int[][] stmax = build(ra); Node[] efs = new Node[80 * n]; int efp = 0; int[][] oas = new int[0][]; for (int i = n - 1; i >= 0; i--) { int[][] noas = new int[40][]; int p = 0; for (int j = 0; j < oas.length; j++) { oas[j][0] |= a[i]; oas[j][1] &= a[i]; if (p > 0 && noas[p - 1][0] == oas[j][0] && noas[p - 1][1] == oas[j][1]) { noas[p - 1][2] = oas[j][2]; } else { noas[p++] = oas[j]; } } if (!(p > 0 && noas[p - 1][0] == a[i] && noas[p - 1][1] == a[i])) { noas[p++] = new int[] { a[i], a[i], i }; } else { noas[p - 1][2] = i; } oas = Arrays.copyOf(noas, p); int to = n; for (int[] oa : oas) { int cha = oa[0] - oa[1]; int low = oa[2] - 1, high = to; while (high - low > 1) { int h = high + low >> 1; if (cha - (-rmq(stmax, i, h + 1) - rmq(stmin, i, h + 1)) >= k) { low = h; } else { high = h; } } if (low >= oa[2]) { efs[efp++] = new Node(i, (long) (low - i + 1) << 32 | i); efs[efp++] = new Node(low + 1, (long) (low - i + 1) << 32 | i); } to = oa[2]; } } Arrays.sort(efs, 0, efp, (efsa, efsb) -> { return efsa.x - efsb.x; }); TreeSet<Long> set = new TreeSet<>(); int q = 0; for (int i = 0; i < n; i++) { int result = -1; while (q < efp && efs[q].x <= i) { long code = efs[q].code; if (set.contains(code)) { set.remove(code); } else { set.add(code); } q++; } if (!set.isEmpty()) { result = Math.max(result, (int) (set.last() >>> 32)); } bw.write(result + "n"); } bw.newLine(); bw.close(); br.close(); } }Problem solution in C++ programming.#include <bits/stdc++.h> using namespace std; #define ms(s, n) memset(s, n, sizeof(s)) #define FOR(i, a, b) for (int i = (a); i < (b); i++) #define FORd(i, a, b) for (int i = (a) - 1; i >= (b); i--) #define FORall(it, a) for (__typeof((a).begin()) it = (a).begin(); it != (a).end(); it++) #define sz(a) int((a).size()) #define pconent(t, x) (t.find(x) != t.end()) #define all(a) (a).begin(), (a).end() #define uni(a) (a).erase(unique(all(a)), (a).end()) #define pb push_back #define pf push_front #define mp make_pair #define fi first #define se second #define prec(n) fixed<<setprecision(n) #define bit(n, i) (((n) >> (i)) & 1) #define bitcount(n) __builtin_popcountll(n) typedef long long ll; typedef unsigned long long ull; typedef long double ld; typedef pair<int, int> pi; typedef vector<int> vi; typedef vector<pi> vii; const int MOD = (int) 1e9 + 7; const int FFTMOD = 1007681537; const int INF = (int) 1e9; const ll LINF = (ll) 1e18; const ld PI = acos((ld) -1); const ld EPS = 1e-9; inline ll gcd(ll a, ll b) {ll r; while (b) {r = a % b; a = b; b = r;} return a;} inline ll lcm(ll a, ll b) {return a / gcd(a, b) * b;} inline ll fpow(ll n, ll k, int p = MOD) {ll r = 1; for (; k; k >>= 1) {if (k & 1) r = r * n % p; n = n * n % p;} return r;} template<class T> inline int chkmin(T& a, const T& val) {return val < a ? a = val, 1 : 0;} template<class T> inline int chkmax(T& a, const T& val) {return a < val ? a = val, 1 : 0;} inline ll isqrt(ll k) {ll r = sqrt(k) + 1; while (r * r > k) r--; return r;} inline ll icbrt(ll k) {ll r = cbrt(k) + 1; while (r * r * r > k) r--; return r;} inline void addmod(int& a, int val, int p = MOD) {if ((a = (a + val)) >= p) a -= p;} inline void submod(int& a, int val, int p = MOD) {if ((a = (a - val)) < 0) a += p;} inline int mult(int a, int b, int p = MOD) {return (ll) a * b % p;} inline int inv(int a, int p = MOD) {return fpow(a, p - 2, p);} inline int sign(ld x) {return x < -EPS ? -1 : x > +EPS;} inline int sign(ld x, ld y) {return sign(x - y);} #define db(x) cerr << #x << " = " << (x) << " "; #define endln cerr << "n"; template<class T> struct RMQOR { int n; vector<T> a; vector<vector<T> > f; T best(T a, T b) { return a | b; } void init(int n) { this->n = n; int p = 1; while ((1 << p) < n) p++; a.resize(n), f.resize(p + 1); for (int i = 0; i < (int) f.size(); i++) { f[i].resize(n); } } void upd(int u, T x) { a[u] = x; } void build() { for (int i = 0; i < n; i++) f[0][i] = a[i]; for (int l = 0, k; (k = 1 << l) < n; l++) { for (int i = 0; i + k < n; i++) { f[l + 1][i] = best(f[l][i], f[l][i + k]); } } } T query(int a, int b) { int l = a == b ? 0 : __lg(b - a); return best(f[l][a], f[l][b - (1 << l) + 1]); } }; RMQOR<long long> rmq_or; template<class T> struct RMQAND { int n; vector<T> a; vector<vector<T> > f; T best(T a, T b) { return a & b; } void init(int n) { this->n = n; int p = 1; while ((1 << p) < n) p++; a.resize(n), f.resize(p + 1); for (int i = 0; i < (int) f.size(); i++) { f[i].resize(n); } } void upd(int u, T x) { a[u] = x; } void build() { for (int i = 0; i < n; i++) f[0][i] = a[i]; for (int l = 0, k; (k = 1 << l) < n; l++) { for (int i = 0; i + k < n; i++) { f[l + 1][i] = best(f[l][i], f[l][i + k]); } } } T query(int a, int b) { int l = a == b ? 0 : __lg(b - a); return best(f[l][a], f[l][b - (1 << l) + 1]); } }; RMQAND<long long> rmq_and; template<class T, class cmp = less<T> > struct RMQ2 { int n; vector<T> a; vector<vector<T> > f; T best(T a, T b) { if (cmp()(a, b)) return a; return b; } void init(int n) { this->n = n; int p = 1; while ((1 << p) < n) p++; a.resize(n), f.resize(p + 1); for (int i = 0; i < (int) f.size(); i++) { f[i].resize(n); } } void upd(int u, T x) { a[u] = x; } void build() { for (int i = 0; i < n; i++) f[0][i] = a[i]; for (int l = 0, k; (k = 1 << l) < n; l++) { for (int i = 0; i + k < n; i++) { f[l + 1][i] = best(f[l][i], f[l][i + k]); } } } T query(int a, int b) { int l = a == b ? 0 : __lg(b - a); return best(f[l][a], f[l][b - (1 << l) + 1]); } }; RMQ2<int> rmq_min; RMQ2<int, greater<int> > rmq_max; const int maxn = 1e5 + 5; const int logn = 31; int n, k; int a[maxn]; int nxt0[maxn][logn]; int nxt1[maxn][logn]; void solve() { cin >> n >> k; rmq_or.init(n); rmq_and.init(n); rmq_min.init(n); rmq_max.init(n); FOR(i, 0, n) { cin >> a[i]; rmq_or.upd(i, a[i]); rmq_and.upd(i, a[i]); rmq_min.upd(i, a[i]); rmq_max.upd(i, a[i]); } rmq_or.build(); rmq_and.build(); rmq_min.build(); rmq_max.build(); static int lst0[logn]; static int lst1[logn]; fill_n(lst0, logn, n); fill_n(lst1, logn, n); FORd(i, n, 0) { FOR(j, 0, logn) { if (!bit(a[i], j)) { lst0[j] = i; } else { lst1[j] = i; } } FOR(j, 0, logn) nxt0[i][j] = lst0[j]; FOR(j, 0, logn) nxt1[i][j] = lst1[j]; } vector<pi> events; FOR(i, 0, n) { int mx = -1; long long cost = 0, sor = a[i], sand = a[i]; int ptr = i; if (cost >= k) { mx = i; } while (ptr < n - 1) { int nptr = n; FOR(j, 0, logn) { if (bit(sand, j)) { chkmin(nptr, nxt0[ptr + 1][j]); } if (!bit(sor, j)) { chkmin(nptr, nxt1[ptr + 1][j]); } } int lo = ptr, hi = nptr - 1; while (lo < hi) { int mi = lo + hi + 1 >> 1; if (rmq_or.query(i, mi) - rmq_and.query(i, mi) - rmq_max.query(i, mi) + rmq_min.query(i, mi) >= k) { lo = mi; } else { hi = mi - 1; } } int mi = lo + hi >> 1; if (rmq_or.query(i, mi) - rmq_and.query(i, mi) - rmq_max.query(i, mi) + rmq_min.query(i, mi) >= k) { mx = mi; } if (nptr == n) { break; } ptr = nptr; sor = rmq_or.query(i, ptr); sand = rmq_and.query(i, ptr); } if (~mx) { int len = mx - i + 1; events.pb(mp(i, len)); events.pb(mp(mx + 1, -len)); } } sort(all(events)); multiset<int> heap; int ptr = 0; FOR(i, 0, n) { while (ptr < sz(events) && events[ptr].fi <= i) { int x = events[ptr].se; if (x > 0) { heap.insert(x); } else { heap.erase(heap.find(-x)); } ptr++; } cout << (sz(heap) ? *heap.rbegin() : -1) << "n"; } } int main(int argc, char* argv[]) { ios_base::sync_with_stdio(0), cin.tie(0); if (argc > 1) { assert(freopen(argv[1], "r", stdin)); } if (argc > 2) { assert(freopen(argv[2], "wb", stdout)); } solve(); cerr << "nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "msn"; return 0; } Problem solution in C programming.#include <stdio.h> #include <stdlib.h> #define INF 200000 int get(int l,int r); int max(int x,int y); int min(int x,int y); void init( int n ); void range_increment( int i, int j, int val ); int query( int i ); void sort_a(int*a,int size,int*new_size); void merge(int*a,int*left,int*right,int left_size, int right_size,int*new_size); int N,a[100000],b[100000],map[100001],dp[4][100000][18],dp1[30][100000],dp2[30][100000],tree[400000]; int main(){ int n,k,s,ns,h,l,m,i,j; scanf("%d%d",&n,&k); for(i=j=1,map[0]=0;i<=100000;i++) if(j*2<=i){ j*=2; map[i]=map[i-1]+1; } else map[i]=map[i-1]; for(i=0;i<n;i++) scanf("%d",a+i); for(i=0;i<n;i++) dp[0][i][0]=dp[1][i][0]=dp[2][i][0]=dp[3][i][0]=a[i]; for(j=1;1<<j<=n;j++) for(i=0;i+(1<<j)-1<n;i++){ dp[0][i][j]=max(dp[0][i][j-1],dp[0][i+(1<<(j-1))][j-1]); dp[1][i][j]=min(dp[1][i][j-1],dp[1][i+(1<<(j-1))][j-1]); dp[2][i][j]=dp[2][i][j-1]|dp[2][i+(1<<(j-1))][j-1]; dp[3][i][j]=dp[3][i][j-1]&dp[3][i+(1<<(j-1))][j-1]; } for(i=0;i<n;i++) for(j=0;j<30;j++) if(a[i]&(1<<j)){ dp1[j][i]=i; dp2[j][i]=INF; } else{ dp1[j][i]=INF; dp2[j][i]=i; } for(i=n-2;i>=0;i--) for(j=0;j<30;j++){ dp1[j][i]=min(dp1[j][i],dp1[j][i+1]); dp2[j][i]=min(dp2[j][i],dp2[j][i+1]); } init(n); for(i=0;i<n;i++){ for(j=s=0;j<30;j++){ if(dp1[j][i]!=INF) b[s++]=dp1[j][i]; if(dp2[j][i]!=INF) b[s++]=dp2[j][i]; } sort_a(b,s,&ns); for(j=ns-1;j>=0;j--) if(get(i,b[j])>=k){ if(j==ns-1) h=n-1; else h=b[j+1]-1; l=s=b[j]; while(l<=h){ m=(h+l)/2; if(get(i,m)>=k){ s=m; l=m+1; } else h=m-1; } range_increment(i,s,s-i+1); break; } } for(i=0;i<n;i++) printf("%dn",query(i)); return 0; } int get(int l,int r){ int a,b,c,d,len; len=map[r-l+1]; a=max(dp[0][l][len],dp[0][r-(1<<len)+1][len]); b=min(dp[1][l][len],dp[1][r-(1<<len)+1][len]); c=dp[2][l][len]|dp[2][r-(1<<len)+1][len]; d=dp[3][l][len]&dp[3][r-(1<<len)+1][len]; return c-d-a+b; } int max(int x,int y){ return (x>y)?x:y; } int min(int x,int y){ return (x<y)?x:y; } void init( int n ) { N = 1; while( N < n ) N *= 2; int i; for( i = 1; i < N + n; i++ ) tree[i] = -1; } void range_increment( int i, int j, int val ) { for( i += N, j += N; i <= j; i = ( i + 1 ) / 2, j = ( j - 1 ) / 2 ) { if( i % 2 == 1 ) tree[i] = max(val,tree[i]); if( j % 2 == 0 ) tree[j] = max(val,tree[j]); } } int query( int i ) { int ans = -1,j; for( j = i + N; j; j /= 2 ) ans = max(tree[j],ans); return ans; } void sort_a(int*a,int size,int*new_size){ if (size < 2){ (*new_size)=size; return; } int m = (size+1)/2,i; int *left,*right; left=(int*)malloc(m*sizeof(int)); right=(int*)malloc((size-m)*sizeof(int)); for(i=0;i<m;i++) left[i]=a[i]; for(i=0;i<size-m;i++) right[i]=a[i+m]; int new_l_size=0,new_r_size=0; sort_a(left,m,&new_l_size); sort_a(right,size-m,&new_r_size); merge(a,left,right,new_l_size,new_r_size,new_size); free(left); free(right); return; } void merge(int*a,int*left,int*right,int left_size, int right_size,int*new_size){ int i = 0, j = 0,index=0; while (i < left_size|| j < right_size) { if (i == left_size) { a[index++] = right[j]; j++; } else if (j == right_size) { a[index++] = left[i]; i++; } else if (left[i] <= right[j]) { a[index++] = left[i]; i++; } else { a[index++] = right[j]; j++; } if(index>1&&a[index-2]==a[index-1]) index--; } (*new_size)=index; return; } coding problems solutions