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Home Code Solutions Hackerrank Algorithms

Arithmetic Expressions – HackerRank Solution

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

admin by admin
August 24, 2022
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Table of Contents

  • Arithmetic Expressions – 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++ replace HackerRank Solution
  • Java rep HackerRank Solution
  • Python 3 rep HackerRank Solution
  • Python 2 rep HackerRank Solution
  • C rep 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:

Arithmetic Expressions – 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++ replace HackerRank Solution


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// Intention is a project that we can change with impunity

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <fstream>
#include <string>
#include <map>
#include <set>
#include <memory>
#include <cstring>
#include <chrono>
#include <climits>


using namespace std;


#ifdef WIN32
#define INPUT my_file
const char* input_file_root = "Input";
const char* input_file_suffix = ".txt";

class OutputCompare
{
public:
	~OutputCompare() {
		if (use_file) {
			if (problems == 0) {
				cout << "Output identical to expected file\n";
			}
			else {
				cout << "Output incorrect, found " << problems << " differences\n";
			}
		}
	}

	void set_file(const char* name) {
		expected_file.open(name);
		use_file = true;
	}

	OutputCompare& operator<<(char c) {
		if (use_file) {
			if (!iswspace(c)) {
				char expected;
				expected_file >> expected;
				if (c != expected) {
					cout << "Expected " << expected << " but program output was " << c << '\n';
					++problems;
				}
			}
		}
		else {
			cout << c;
		}
		return *this;
	}

	template <class T>
	OutputCompare& operator<<(const T& value) {
		if (use_file) {
			T expected;
			expected_file >> expected;
			if (value != expected) {
				cout << "Expected " << expected << " but program output was " << value << '\n';
				++problems;
			}
		}
		else {
			cout << value;
		}
		return *this;
	}

private:
	ifstream expected_file;
	int problems = 0;
	bool use_file = false;
};
#define OUTPUT my_output_compare

#else
#define INPUT cin
#define OUTPUT cout
#define _ASSERT(x)
#endif

int main(int argc, const char * argv[])
{
	// Read in problem
#ifdef WIN32
	// Read from a file.  Command line argument [1] is appended to the file name.
	ifstream my_file;
	string input_file_name(input_file_root);
	if (argc >= 2) {
		input_file_name.append(argv[1]);
	}
	input_file_name.append(input_file_suffix);
	my_file.open(input_file_name);

	OutputCompare my_output_compare;
	if (argc < 3 || *argv[2] != 'x') {
		string output_file_name("Output");
		if (argc >= 3) {
			output_file_name.append(argv[2]);
		}
		else if (argc >= 2) {
			output_file_name.append(argv[1]);
		}
		output_file_name.append(".txt");
		my_output_compare.set_file(output_file_name.c_str());
	}
	auto start_time = std::chrono::system_clock::now();
#endif

	int n;
	INPUT >> n;
	int* data = new int[n];
	char* operators = new char[n * 101];
	char* follows = new char[n * 101];

	for (int i = 0; i < n; ++i) {
		INPUT >> data[i];

		for (int j = 0; j < 101; ++j) {
			operators[i * 101 + j] = ' ';
		}
	}

	// Specify that the first input number can be reached
	operators[data[0]] = '+';

	for (int i = 1; i < n; ++i) {
		char* last_ops = operators + (i - 1) * 101;
		char* this_ops = operators + i * 101;
		char* this_follows = follows + i * 101;

		// See what numbers can be reached by combining numbers reachable after i-1 steps with data[i]
		// Note that once we've found a solution, multiplying each later operand will give correct answer
		// So we shortcircuit once entry 0 is filled in
		for (int j = 0; j < 101 && this_ops[0] == ' '; ++j) {
			if (last_ops[j] != ' ') {
				int reachable = j + data[i];
				reachable %= 101;
				if (this_ops[reachable] == ' ') {
					this_ops[reachable] = '+';
					this_follows[reachable] = j;
				}

				reachable = j - data[i];
				if (reachable < 0) reachable += 101;
				if (this_ops[reachable] == ' ') {
					this_ops[reachable] = '-';
					this_follows[reachable] = j;
				}

				reachable = j * data[i];
				reachable %= 101;
				if (this_ops[reachable] == ' ') {
					this_ops[reachable] = '*';
					this_follows[reachable] = j;
				}
			}
		}
	}

	// We should have a result (i.e. 0 should be included in last one)
	if (operators[(n - 1) * 101] == ' ') {
		OUTPUT << "ERROR!!!!!\n";
	}
	else {
		// We obtain the result in reverse order
		vector<char> result;
		result.reserve(n);
		int x = 0;
		for (int i = n - 1; i > 0; --i) {
			result.push_back(operators[i * 101 + x]);
			x = follows[i * 101 + x];
		}
		_ASSERT(x == data[0]);
		OUTPUT << data[0];
		auto it = result.rbegin();
		for (int i = 1; i < n; ++i, ++it) {
			OUTPUT << (*it);
			OUTPUT << data[i];
		}
		OUTPUT << '\n';
	}


	return 0;

#ifdef WIN32
	my_file.close();
	cout << "time=" << std::chrono::duration_cast<std::chrono::milliseconds>(std::chrono::system_clock::now() - start_time).count() << "ms\n";
#endif
	return 0;
}

Java rep HackerRank Solution


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import java.io.*;
import java.util.*;

public class Solution {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] a = new int[n];
        for(int i = 0; i < n; i++) {
            a[i] = sc.nextInt();            
        }
        
        LinkedList<Character> sol = new LinkedList<Character>();
        int[] inv = new int[101];
        inv[1] = 1;
        for(int i = 2; i <= 100; i++) {
            for(int j=i; j <= 100; j++) {
                if(((i*j)%101) == 1) {
                    inv[i] = j;
                    inv[j] = i;
                }
            }
        }
        
        backtrack(a,n-1,0,sol,inv);
        
        while(sol.size() < n-1) {
            sol.addFirst('*');
        }
        
        for(int i = 0; i < n-1; i++) {
            System.out.print(a[i]+""+sol.get(i));
        }
        System.out.println(a[n-1]);
        
        
    }
    
    public static boolean backtrack(int[] a, int index, int mod, LinkedList<Character>sol, int[] inv) {
        if(index == 0) {
            return ((a[index] % 101) == mod);
        }
        
        if(a[index] == 0) {
            if(mod == 0) {
                sol.add('*');
                return true;
            }
        } else {
            if(mod == 0) {
                if(backtrack(a,index-1,0,sol,inv)) {
                    sol.add('*');
                    return true;
                }
            } else {
                if(backtrack(a,index-1,(mod*inv[a[index]])%101,sol,inv)) {
                    sol.add('*');
                    return true;
                }
            }
        }
        
        if(backtrack(a,index-1,(mod+a[index])%101,sol,inv)) {
            sol.add('-');
            return true;
        }
        
        int t = mod-a[index];
        if(t < 0) {
            t = t + 101;
        }
        if(backtrack(a,index-1,t,sol,inv)) {
            sol.add('+');
            return true;
        }
        
        return false;
    }
    
    
}



Python 3 rep HackerRank Solution


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import sys

sys.setrecursionlimit(15000)

ops = [lambda x,y: x + y, lambda x,y: x * y, lambda x,y: x - y]
op2s = ['+', '*', '-', '']

def exp(i, value, l):
    if i == n:
        return l if value % 101 == 0 else None
    
    if len(cache[i]) > 0 and value in cache[i]:
        return cache[i][value]
    
    for k in range(3):
        if exp(i + 1, ops[k](value, a[i]), l) != None:
            l[i - 1] = k
            cache[i][value] = l
            return l
        
    cache[i][value] = None
    return None

n = int(input())
a = [int(x) for x in input().strip().split(' ')]

cache = [{}] * n

l = exp(1, a[0], [3] * n)

for i in range(n):
    sys.stdout.write(str(a[i]))
    sys.stdout.write(op2s[l[i]])
    
sys.stdout.flush()



Python 2 rep HackerRank Solution


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# Enter your code here. Read input from STDIN. Print output to STDOUT
import itertools

N = int(raw_input().strip())
nums = map(int, raw_input().split())
seqs = itertools.product('*+-', repeat=N-1)

def get_seq(ops, n):
    for op in ops:
        x = n
        for s in range(len(op)):
            if op[s]=='+':
                x = x + nums[s+1]
            elif op[s]=='-':
                x = x - nums[s+1]
            else:
                x = x * nums[s+1]
        if x % 101 == 0:
            return op

seq = get_seq(seqs,nums[0])

print ''.join([str(nums[0])]+[seq[i]+str(nums[i+1]) for i in range(len(seq))])



C rep HackerRank Solution


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#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
  int n;
  scanf("%d", &n);

  unsigned char * ops = malloc(n);  // '+' - 1, '-' - 2, '*' - 3
  memset(ops, 3, n);

  int * nums = malloc(n * sizeof(unsigned int));
  for (int i = 0; i < n; i ++) {
    scanf("%d", &nums[i]);
  }

  unsigned char t[1000][101][2];

  int i;
  for (i = 0; i < n; i ++) {
    if (0 == i) {
      t[i][nums[i]][0] = 1;
      t[i][nums[i]][1] = 1;
    }
    else {
      int j;
      for (j = 1; j < 101; j ++) {
        if (t[i - 1][j][0]) {
          int tmp = (j + nums[i]) % 101;
          t[i][tmp][0] = j;
          t[i][tmp][1] = 1;
          if (0 == tmp) {
            break;
          }

          tmp = (101 + j - nums[i]) % 101;
          t[i][tmp][0] = j;
          t[i][tmp][1] = 2;
          if (0 == tmp) {
            break;
          }

          tmp = (j * nums[i]) % 101;
          t[i][tmp][0] = j;
          t[i][tmp][1] = 3;
          if (0 == tmp) {
            break;
          }
        }
      }

      if (j < 101) {
        break;
      }
    }
  }

  int p = 0;
  for (int j = i; j > 0; j --) {
    ops[j] = t[j][p][1];
    p = t[j][p][0];
  }

  for (int i = 0; i < n - 1; i ++) {
    printf("%d", nums[i]);
    switch (ops[i + 1]) {
      case 1:
      printf("+");
      break;
      case 2:
      printf("-");
      break;
      default:
      printf("*");
      break;
    }
  }
  printf("%d\n", nums[n - 1]);

  free(nums);
  free(ops);

  return 0;
}

 

Warmup
Implementation
Strings
Sorting
Search
Graph Theory
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Dynamic Programming
Constructive Algorithms
Bit Manipulation
Recursion
Game Theory
NP Complete
Debugging

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