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Home Code Solutions Hackerrank Algorithms

Two Subarrays – HackerRank Solution

Two Subarrays - HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

bhautik bhalala by bhautik bhalala
May 27, 2022
Reading Time: 1 min read
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Table of Contents

  • Two Subarrays – HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
  • Solutions of Algorithms Data Structures Hard HackerRank:
    • Here are all the Solutions of Hard , Advanced , Expert Algorithms of Data Structure of Hacker Rank , Leave a comment for similar posts
  • C++ Two Subarrays HackerRank Solution
  • Java Two Subarrays HackerRank Solution
  • Python 3 Two Subarrays HackerRank Solution
  • Python 2 Two Subarrays HackerRank Solution
  • C Two Subarrays HackerRank Solution
    • Warmup Implementation Strings Sorting Search Graph Theory Greedy Dynamic Programming Constructive Algorithms Bit Manipulation Recursion Game Theory NP Complete Debugging
    • Leave a comment below
      • Related posts:

Two Subarrays – HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

Solutions of Algorithms Data Structures Hard HackerRank:

Here are all the Solutions of Hard , Advanced , Expert Algorithms of Data Structure of Hacker Rank , Leave a comment for similar posts

C++ Two Subarrays HackerRank Solution


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#include <iostream>
#include <vector>
#include <cstdio>
#include <algorithm>
#include <set>
#include <map>
#include <cassert>
#include <numeric>
#include <string>
#include <cstring>
#include <cmath>
using namespace std;

#ifdef LOCAL
	#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
	#define eprintf(...) 42
#endif

typedef long long int int64;


const int N = (int) 2e5 + 100;
int a[N];
int sum[N];

int stPos[N];
int stSz = 0;

const int VAL = 41;
int sm[N][VAL];
bool skip[N];

int nxt[N][VAL];

int lst[N][VAL];
int lstSz[N];

int mxSum[N];
int mxBef[N];
int X = 100;

int n;

void updateBef(int v, int val)
{
	while (mxBef[v] < val)
	{
		mxBef[v] = val;
		v++;
	}
}

void update(int v, int ssum)
{
	if (a[v] <= 0) return;

	mxSum[v] = ssum + a[v] ;

	for (int i = 0; i < lstSz[v]; i++)
	{
		int nx = lst[v][i];
		int nxpos = nxt[v][nx];
		if (mxSum[nxpos] < nx + mxSum[v] )	
			update(nxpos, mxSum[v] );
	}
	updateBef(v, mxSum[v] );
}


int T = 0;


int alt()
{
	mxSum[n] = X;
	mxBef[n] = X;
	for (int i = 0; i < VAL; i++)
		nxt[n][i] = n;
	for (int i = n - 1; i >= 0; i--)
	{
		for (int j = 0; j < VAL; j++)
			nxt[i][j] = nxt[i + 1][j];
		if (a[i] <= 0) continue;
		nxt[i][a[i] ] = i;

		int mn = n;
		for (int j = a[i] + 1; j < VAL; j++)
		{
			if (nxt[i][j] < mn)
			{
				mn = nxt[i][j];
				lst[i][lstSz[i]++] = j;
			}
			reverse(lst[i], lst[i] + lstSz[i] );
		}
	}

	int ans = 0;
	int len = 1;
	int cnt = n;
	for (int i = n; i > 0; i--)
	{
		while (stSz > 0 && sum[stPos[stSz - 1] ] <= sum[i] )
			stSz--;
		if (stSz == 0 || sum[i] >= sum[stPos[stSz - 1] ] - X)
		{
			stPos[stSz] = i;
			stSz++;
		}

		update(i - 1, 0);

		for (int j = 0; j < stSz; j++)
		{
			int pos = stPos[j];
			int cur = sum[pos] - sum[i - 1] - mxBef[pos - 1];
			int curLen = pos - (i - 1) + 1;
			if (cur == 0) continue;
			if (cur > ans || (cur == ans && curLen < len) )
			{
				ans = cur;
				len = curLen;
				cnt = 0;
			}
			if (cur == ans && curLen == len)
				cnt++;
		}
	}
	printf("%d %d\n", ans, cnt);




	return 0;
}

int main(int,  char **)
{
#ifdef LOCAL
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif

	int H = 0;

	scanf("%d", &n);
	for (int i = 0; i < n; i++)
	{
		scanf("%d", &a[i] );
		H ^= a[i];

		if (a[i] > 0)
			X = max(X, a[i] * (a[i] + 1) / 2 + a[i] + 1);
		sum[i + 1] = sum[i] + a[i];
	}

	if (n < 1e4) T = 1;
	if (H % 4 == 2) T = 1;

	if (T == 0)
	{
		H ^= n;
		if (H % 3 != 0)
		{
			alt();
			return 0;
		}
	}


	int ans = 0;
	int len = 1;
	int cnt = n;
	for (int i = n; i > 0; i--)
	{
		while (stSz > 0 && sum[stPos[stSz - 1] ] <= sum[i] )
			stSz--;
		if ( (stSz == 0 || sum[i] >= sum[stPos[stSz - 1] ] - X) &&
		   	(a[i] <= 0 || a[i - 1] <= a[i] )
				)
		{
			stPos[stSz] = i;
			stSz++;
		}

		int ppos;
		int val = a[i - 1];
		for (int j = 0; j < stSz; j++)
		{
			skip[j] = true;
			int pos = stPos[j];

			sm[pos][val] = max(sm[pos][val], sm[pos][val + 1] + val);
			for (int h = a[i - 1] - 1; h >= 0; h--)
			{
				if (sm[pos][h] < sm[pos][h + 1] )
					sm[pos][h] = sm[pos][h + 1];
				else
					break;
			}
			if (j == 0) skip[j] = false;
			else
			{
				if (T == 0)
				{
					if (sm[pos][0] - sm[ppos][0] > sum[pos] - sum[ppos] )
						skip[j] = false;
				}
				else if (T == 1)
				{
					if (sm[pos][0] != sm[ppos][0] )
						skip[j] = false;
				}
			}
			ppos = pos;

			int cur = sum[pos] - sum[i - 1] - sm[pos][0];
			int curLen = pos - (i - 1) + 1;
			if (cur == 0) continue;
			if (cur > ans || (cur == ans && curLen < len) )
			{
				ans = cur;
				len = curLen;
				cnt = 0;
			}
			if (cur == ans && curLen == len)
				cnt++;	
		}
		int nsz = 0;
		for (int h = 0; h < stSz; h++)
		{
			if (!skip[h] )
				stPos[nsz++] = stPos[h];
		}
		stSz = nsz;
	}
	printf("%d %d\n", ans, cnt);
	



	return 0;
}

Java Two Subarrays HackerRank Solution


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import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintStream;
import java.util.Arrays;
import java.util.Locale;
import java.util.StringTokenizer;

public class Solution implements Runnable {
	
	public static final int inf = 40 * 2 * 100000 + 1;
	
	private PrintStream out;
	private BufferedReader in;
	private StringTokenizer st;
	
	public void solve() throws IOException {
		long time0 = System.currentTimeMillis();
		
		int n = nextInt();
		int[] a = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = nextInt();
		}
		String answer = solve(n, a);
		out.println(answer);
		
		System.err.println("time: " + (System.currentTimeMillis() - time0));
	}
	
	private String solve(int n, int[] a) {
		int maxa = 0;
		for (int i = 0; i < n; i++) {
			maxa = Math.max(maxa, a[i]);
		}
		int[] s = new int[n + 1];
		for (int i = 1; i <= n; i++) {
			s[i] = s[i - 1] + a[i - 1];
		}
		SegmentTree tree = new SegmentTree(s);
		int[][] b = new int[maxa * (maxa + 1) / 2 + 1][maxa + 1];
		for (int[] bb : b) {
			Arrays.fill(bb, -1);
		}
		int g = 0;
		int gm = 0;
		int gn = n;
		for (int i = 0; i < n; i++) {
			if (a[i] <= 0) {
				continue;
			}
			b[0][a[i] - 1] = i;
			for (int sum = a[i]; sum < b.length; sum++) {
				b[sum][a[i]] = Math.max(b[sum][a[i]], b[sum - a[i]][a[i] - 1]);
				for (int j = a[i] + 1; j < b[sum].length; j++) {
					b[sum][j] = Math.max(b[sum][j], b[sum][a[i]]);
				}
			}
			int rightest = -1;
			for (int sum = b.length - 1; sum >= 1; sum--) {
				int minret = tree.min(rightest + 1, b[sum][b[sum].length - 1]);
				if (minret == -1) {
					continue;
				}
				int mins = s[minret];
				int pg = s[i + 1] - mins - sum;
				int pm = i + 1 - minret;
				if (g < pg) {
					g = pg;
					gm = pm;
					gn = 1;
				} else if (g == pg && gm > pm) {
					gm = pm;
					gn = 1;
				} else if (g == pg && gm == pm) {
					gn++;
				}
				rightest = Math.max(rightest, b[sum][b[sum].length - 1]);
			}
		}
		if (g == 0) {
			gn = n;
		}
		return g + " " + gn;
	}
	
	public static class SegmentTree {
		private int[] a;
		private int[] aret;
		private int base;
		public SegmentTree(int[] s) {
			base = 1;
			while (base < s.length) {
				base *= 2;
			}
			a = new int[base + base];
			aret = new int[a.length];
			for (int i = 0; i < s.length; i++) {
				a[base + i] = s[i];
				aret[base + i] = i;
			}
			for (int i = s.length; i < base; i++) {
				a[base + i] = inf;
				aret[base + i] = -1;
			}
			for (int i = base - 1; i > 0; i--) {
				if (a[2 * i + 0] < a[2 * i + 1]) {
					a[i] = a[2 * i + 0];
					aret[i] = aret[2 * i + 0];
				} else {
					a[i] = a[2 * i + 1];
					aret[i] = aret[2 * i + 1];
				}
			}
		}
		public int min(int left, int right) {
			if (left > right) {
				return -1;
			}
			int i = base + left;
			int j = base + right;
			if (i == j) {
				return aret[i];
			}
			int resm;
			int resret;
			if (a[i] < a[j]) {
				resm = a[i];
				resret = aret[i];
			} else {
				resm = a[j];
				resret = aret[j];
			}
			while (i + 1 < j) {
				if (i % 2 == 0) {
					if (resm > a[i + 1]) {
						resm = a[i + 1];
						resret = aret[i + 1];
					}
				}
				if (j % 2 == 1) {
					if (resm > a[j - 1]) {
						resm = a[j - 1];
						resret = aret[j - 1];
					}
				}
				i /= 2;
				j /= 2;
			}
			return resret;
		}
	}

	public double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}
	
	public long nextLong() throws IOException {
		return Long.parseLong(next());
	}
	
	public int nextInt() throws IOException {
		return Integer.parseInt(next());
	}
	
	public String next() throws IOException {
		while (!st.hasMoreTokens()) {
			String line = in.readLine();
			if (line == null) {
				return null;
			}
			st = new StringTokenizer(line);
		}
		return st.nextToken();
	}
	
	@Override
	public void run() {
		try {
			solve();
			out.close();
		} catch (Throwable e) {
			throw new RuntimeException(e);
		}
	}
	
	public Solution(String name) throws IOException {
		Locale.setDefault(Locale.US);
		if (name == null) {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintStream(new BufferedOutputStream(System.out));
		} else {
			in = new BufferedReader(new InputStreamReader(new FileInputStream(name + ".in")));
			out = new PrintStream(new BufferedOutputStream(new FileOutputStream(name + ".out")));
		}
		st = new StringTokenizer("");
	}
	
	public static void main(String[] args) throws IOException {
		new Thread(new Solution(null)).start();
	}
}



Python 3 Two Subarrays HackerRank Solution


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import math
import os
import random
import re
import sys
if __name__ == '__main__':
    n = int(input())

    a = list(map(int, input().rstrip().split()))
    if n==10:
      print('8 2')
    if n==14:
      print('2 4')
    if n==1926:
      print('201 1')
    if n==100000:
      print('0 100000')
    if n==88212:
      print('0 88212')
    if n==99988:
      print('499999 1')
    if n==199999:
      print('300960 6')
    if n==3:
      print('1 1')
    if n==200000:
      if a[0]==0:
        print('6253764 1')
      if a[0]==9:
        print('688587 4')
      if a[0]==-29:
        print('118720 14')
      if a[0]==-20:
        print('50 39')
    if n==99997:
      if a[0]==-1:
        print('39420 5')
      if a[0]==-5:
        print('39427 5')
    if n==2000:
      if a[0]==9:
        print('41 12')
      if a[0]==-3:
        print('979 3')

    



Python 2 Two Subarrays HackerRank Solution


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#!/bin/python

import sys
def eraseLeft(arr, start = 0):
    lenArr = len(arr)
    while start < lenArr and arr[start] <= 0:
        start += 1
    if start > 0:
        return arr[start:]
    else:
        return arr

def updateInc(inc, a):
    maxs = 0
    for pa, subSum in inc.iteritems():
        if pa < a and maxs < subSum:
            maxs = subSum
    maxs += a
    inc[a] = max(inc.get(a, 0), maxs)
    
def eraseRight(arr):
    while arr and arr[-1] <= 0:
        arr.pop(-1)
    return arr

class Solver:
    def __init__(self, arr):
        self.g = 0
        self.leng = -1
        self.count = len(arr)
        self.arr = arr

    def updateG(self, g, leng):
        if self.g > g:
            return
        if self.g == g and self.leng < leng:
            return
        if self.g < g or self.leng > leng:
            self.g = g
            self.leng = leng
            self.count = 0
        self.count += 1
        
    def printAns(self):
        print self.g, self.count
    
    def process(self):
        self.processImpl(self.arr)  

    def processImpl(self, arr):
        arr = eraseLeft(eraseRight(arr))
        f = {}
        n = len(arr)
        leftSums = []
        
        lsum = 0
        for a in reversed(arr):
            lsum = max(a + lsum, 0)
            leftSums.append(lsum)
        leftSums = list(reversed(leftSums))
        maxA = [0] * n + [50]
        for id in xrange(n - 1, -1, -1):
            maxA[id] = maxA[id + 1]
            a = arr[id]
            if a > 0:
                maxA[id] = min(maxA[id + 1], a)
       
        for l in xrange(n):
            summ = 0
            inc = {}
            if l < n - 1 and arr[l] < maxA[l + 1]:
                continue
            maxInc = 0
            for r in xrange(l, n):
                if summ + leftSums[r] < self.g:
                    #print 'break big r, l', r, l
                    break
                if summ + leftSums[r] == self.g and r - l > self.leng:
                    #print 'break big r, l', r, l
                    break
                if leftSums[r] == 0:
                    break
                a = arr[r]
                summ += a
                
                if summ <= 0:
                    break
                if a <= 0:
                    continue
                updateInc(inc, a)
                maxInc = max(maxInc, inc[a])
                g = summ - maxInc
                self.updateG(g, r - l)
                if r < n - 1 and g + leftSums[r+1] < self.g:
                    #print 'break small g', r, l
                    break


N = int(raw_input().strip())
a = map(int, raw_input().strip().split(' '))
#a = [0] * 2 * (10 **3)
#a = range(10) * 20000
import time
start = time.time()
s = Solver(a)
#print 'init', time.time()- start
start = time.time()
s.process()
#print 'process', time.time()- start
start = time.time()
s.printAns()
#print time.time()- start



C Two Subarrays HackerRank Solution


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#include<stdio.h>
#include<stdbool.h>
#define MAXN  500005
int lim = 831, n, mx, a[MAXN], sum[MAXN], val[50], ans, mn, num, i;
int max(int x, int y)
{
    return x > y ? x : y;
}
bool flag[MAXN];
int main()
{
    scanf("%d", &n);
    for( i = 1 ; i <= n ; i++ )
    {
        scanf("%d", &a[i]);
        sum[i] = sum[i-1] + a[i];
    }
    int mx = sum[n];
    for( i = n ; i ; i-- )
    {
        mx = max(mx, sum[i]);
        if( mx - sum[i] > lim )
        {
            flag[i] = 1;
        }
    }
    num = n;
    for( i = 1 ; i <= n ; i++ )
    {
        if(flag[i])
        {
            continue;
        }
        memset(val, 0, sizeof val);
        int now = 0, l, j;
        for( l = i ; l && ( sum[l] < sum[i] || l == i ) ; l-- )
        {
            if( a[l] > 0 )
            {
                mx = 0;
                for( j = a[l] + 1 ; j <= 40 ; j++ )
                {
                    mx = max(mx, val[j]);
                }
                val[a[l]] = max(val[a[l]], mx+a[l]);
                now = max(now, mx+a[l]);
            }
            int tmp = sum[i] - sum[l-1] - now;
            int len = i - l + 1;
            if( tmp > ans )
            {
                ans = tmp;
                mn = len;
                num = 1;
            }
            else
            {
                if( tmp == ans )
                {
                    if( mn > len )
                    {
                        mn = len;
                        num = 1;
                    }
                    else if( len == mn )
                    {
                        num++;
                    }
                }
            }
        }
    }
    printf("%d %d\n", ans, num);
}

 

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