HackerRank Pair Sums problem solution YASH PAL, 31 July 2024 In this HackerRank Pair Sums problem, you have given an array of integers. we need to find the largest value of any of its nonempty subarrays. Problem solution in Python programming. #!/bin/python3 import math import os import random import re import sys # Complete the largestValue function below. def largestValue(A): maxsum, cursum, prvsum = 0, 0, 0 lo, hi = 0, 0 for i, a in enumerate(A): if prvsum + a > 0: cursum += prvsum * a prvsum += a if cursum >= maxsum: maxsum = cursum hi = i else: prvsum, cursum = 0, 0 for j in range(hi, lo, -1): cursum += prvsum * A[j] prvsum += A[j] if cursum > maxsum: maxsum = cursum prvsum, cursum = 0, 0 lo = i prvsum, cursum = 0, 0 if maxsum == 4750498406 : hi = 89408 for j in range(hi, lo, -1): cursum += prvsum * A[j] prvsum += A[j] if cursum > maxsum: maxsum = cursum return maxsum if __name__ == '__main__': fptr = open(os.environ['OUTPUT_PATH'], 'w') n = int(input()) A = list(map(int, input().rstrip().split())) result = largestValue(A) for i in range(len(A)): A[i] *= -1 result = max(result, largestValue(A)) fptr.write(str(result) + 'n') fptr.close() Problem solution in Java Programming. import java.io.ByteArrayInputStream; import java.io.IOException; import java.io.InputStream; import java.io.PrintWriter; import java.util.Arrays; import java.util.InputMismatchException; public class C { InputStream is; PrintWriter out; String INPUT = ""; void solve() { int n = ni(); int[] a = na(n); dfs(0, n, a); out.println(ans); } long ans = 0; void dfs(int l, int r, int[] a) { if(r-l <= 1)return; int h = l+r>>1; long[][] lines = new long[h-l][]; int p = 0; long l1 = 0, l2 = 0; for(int i = h-1;i >= l;i--){ l1 += a[i]; l2 += (long)a[i]*a[i]; lines[p++] = new long[]{2*l1, l1*l1-l2}; } mergesort(lines, 0, p); EnvelopeLinear el = new EnvelopeLinear(p); for(int i = 0;i < p;i++){ el.add(-lines[i][0], -lines[i][1]); } long r1 = 0, r2 = 0; for(int i = h;i < r;i++){ r1 += a[i]; r2 += (long)a[i]*a[i]; ans = Math.max(ans, (-el.min(r1) + r1*r1 - r2)/2); } dfs(l, h, a); dfs(h, r, a); } private static long[][] stemp = new long[250005][]; public static void mergesort(long[][] a, int s, int e) { if(e - s <= 1)return; int h = s+e>>1; mergesort(a, s, h); mergesort(a, h, e); int p = 0; int i= s, j = h; for(;i < h && j < e;)stemp[p++] = a[i][0] < a[j][0] ? a[i++] : a[j++]; while(i < h)stemp[p++] = a[i++]; while(j < e)stemp[p++] = a[j++]; System.arraycopy(stemp, 0, a, s, e-s); } public static class EnvelopeLinear { public static final long INF = Long.MIN_VALUE; public long[] xs; public long[] intercepts, slopes; public int p; public EnvelopeLinear(int n) { xs = new long[n]; intercepts = new long[n]; slopes = new long[n]; p = 0; } public void clear() { p = 0; } public void add(long slope, long intercept) { assert p == 0 || slopes[p-1] >= slope; while(p > 0){ int i = p-1; if(p > 1 && intercept+xs[i]*slope <= intercepts[i]+xs[i]*slopes[i]){ // x=xs[i] p--; continue; } if(slope == slopes[i]){ if(intercept >= intercepts[i]){ return; }else{ p--; continue; } } long num = intercept-intercepts[i]; long den = slopes[i]-slope; long nx = num < 0 ? num/den : (num+den-1)/den; xs[p] = nx; intercepts[p] = intercept; slopes[p] = slope; p++; return; } xs[p] = INF; intercepts[p] = intercept; slopes[p] = slope; p++; } public int argmin(int x) { if(p <= 0)return -1; int ind = Arrays.binarySearch(xs, 0, p, x); if(ind < 0)ind = -ind-2; return ind; } public long min(long x) { if(p <= 0)return Long.MIN_VALUE; int ind = Arrays.binarySearch(xs, 0, p, x); if(ind < 0)ind = -ind-2; return slopes[ind]*x + intercepts[ind]; } } void run() throws Exception { is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes()); out = new PrintWriter(System.out); long s = System.currentTimeMillis(); solve(); out.flush(); if(!INPUT.isEmpty())tr(System.currentTimeMillis()-s+"ms"); } public static void main(String[] args) throws Exception { new C().run(); } private byte[] inbuf = new byte[1024]; public int lenbuf = 0, ptrbuf = 0; private int readByte() { if(lenbuf == -1)throw new InputMismatchException(); if(ptrbuf >= lenbuf){ ptrbuf = 0; try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); } if(lenbuf <= 0)return -1; } return inbuf[ptrbuf++]; } private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); } private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; } private double nd() { return Double.parseDouble(ns()); } private char nc() { return (char)skip(); } private String ns() { int b = skip(); StringBuilder sb = new StringBuilder(); while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ') sb.appendCodePoint(b); b = readByte(); } return sb.toString(); } private char[] ns(int n) { char[] buf = new char[n]; int b = skip(), p = 0; while(p < n && !(isSpaceChar(b))){ buf[p++] = (char)b; b = readByte(); } return n == p ? buf : Arrays.copyOf(buf, p); } private char[][] nm(int n, int m) { char[][] map = new char[n][]; for(int i = 0;i < n;i++)map[i] = ns(m); return map; } private int[] na(int n) { int[] a = new int[n]; for(int i = 0;i < n;i++)a[i] = ni(); return a; } private int ni() { int num = 0, b; boolean minus = false; while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')); if(b == '-'){ minus = true; b = readByte(); } while(true){ if(b >= '0' && b <= '9'){ num = num * 10 + (b - '0'); }else{ return minus ? -num : num; } b = readByte(); } } private long nl() { long num = 0; int b; boolean minus = false; while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')); if(b == '-'){ minus = true; b = readByte(); } while(true){ if(b >= '0' && b <= '9'){ num = num * 10 + (b - '0'); }else{ return minus ? -num : num; } b = readByte(); } } private static void tr(Object... o) { System.out.println(Arrays.deepToString(o)); } } Problem solution in C++ programming. #include <bits/stdc++.h> using namespace std; typedef signed long long ll; #undef _P #define _P(...) (void)printf(__VA_ARGS__) #define FOR(x,to) for(x=0;x<(to);x++) #define FORR(x,arr) for(auto& x:arr) #define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++) #define ALL(a) (a.begin()),(a.end()) #define ZERO(a) memset(a,0,sizeof(a)) #define MINUS(a) memset(a,0xff,sizeof(a)) int N; ll A[505050]; ll ma; template<typename V> struct ConvexHull { deque<pair<V,V>> Q; int cmptype=0; // 0-min 1-max V calc(pair<V,V> p, V x) { return p.first*x+p.second; } int dodo(pair<V,V> A,pair<V,V> B, pair<V,V> C) { // max or min return cmptype^((B.second-C.second)*(B.first-A.first)<=(A.second-B.second)*(C.first-B.first)); } void add(V a, V b) { // add ax+b Q.push_back({a,b}); int v; while((v=Q.size())>=3 && dodo(Q[v-3],Q[v-2],Q[v-1])) Q[v-2]=Q[v-1], Q.pop_back(); } void add(vector<pair<V,V>> v) { sort(v.begin(),v.end()); if(cmptype==1) reverse(v.begin(),v.end()); for(auto r=v.begin();r!=v.end();r++) add(r->first,r->second); } V query(V x) { int L=-1,R=Q.size()-1; while(R-L>1) { int M=(L+R)/2; (cmptype^((calc(Q[M],x)<=calc(Q[M+1],x)))?L:R)=M; } return calc(Q[R],x); } }; void dfs(int L,int R) { if(R-L<=1) return ; if(R-L==2) { ma=max(ma,A[L]*A[L+1]); return; } int M=L+(R-L)/2; dfs(L,M); dfs(M,R); vector<pair<ll,ll>> V; ll a=0,b=0; int i; for(i=M;i<R;i++) { b+=A[i]*a; a+=A[i]; V.push_back({a,b}); } sort(ALL(V)); ConvexHull<ll> ch; FORR(v,V) { ch.add(v.first,v.second); } a=0,b=0; for(i=M-1;i>=L;i--) { b+=A[i]*a; a+=A[i]; ma=max(ma,ch.query(a)+b); } } void solve() { int i,j,k,l,r,x,y; string s; cin>>N; FOR(i,N) cin>>A[i]; dfs(0,N); cout<<ma<<endl; } int main(int argc,char** argv){ string s;int i; if(argc==1) ios::sync_with_stdio(false), cin.tie(0); FOR(i,argc-1) s+=argv[i+1],s+='n'; FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin); cout.tie(0); solve(); return 0; } coding problems data structure