[ACCEPTED]-Project Euler #8, I don't understand where I'm going wrong-c++

Score: 27

In fact your solution is too small rather 23 than too big. The answer is what was pointed 22 out in the comments, that there is integer 21 overflow, and the clue is in the fact that 20 your solution is close to the largest possible 19 value for an signed int: 2147483647. You 18 need to use a different type to store the 17 product.

Note that the answer below is still 16 'correct' in that your code does do this 15 wrong, but it's not what is causing the 14 wrong value. Try taking your (working) code 13 to http://codereview.stackexchange.com if you would like the folks there to 12 tell you what you could improve in your 11 approach and your coding style.

You are checking 10 for a new greatest product inside the inner 9 loop instead of outside. This means that 8 your maximum includes all strings of less 7 or equal ton 13 digits, rather than only 6 exactly 13.

This could make a difference 5 if you are finding a string which has fewer 4 than 13 digits which have a large product, but 3 a 0 at either end. You shouldn't count this 2 as the largest, but your code does. (I haven't 1 checked if this does actually happen.)

``````for (int i=0; i < num.length() -12; i++)
{
product = ((int) num[i] - 48);
for (int j=i+1; j<i+13; j++)
{
product = product * ((int) num[j] - 48);
}
if (greatestProduct <= product)
{
greatestProduct = product;
}
}
``````
Score: 6

9^13 ≈ 2.54e12 (maximal possible value, needs 2 42 bit to be represented exactly), which 1 doesn't fit into `signed int`. You should use int64.

Score: 2

If you don't want to mess with BigNum libraries, you 3 could just take logarithms of your digits 2 (rejecting 0) and add them up. It amounts 1 to the same comparison.

Score: 1

A faster way without internal loop, but 3 works only where there isn't a `0` in the input:

``````long long greatest(string num)
{
if (num.length() < 13)
return 0;

// Find a product of first 13 numbers.
long long product = 1;
unsigned int i;
for (i=0; i<13; ++i) {
product *= (num[i]-'0');
}
long long greatest_product = product;
// move through the large number
for (i=0; i+13<num.length(); ++i) {
product = product/(num[i]-'0')*(num[i+13]-'0');
if (greatest_product < product)
greatest_product = product;
}
return greatest_product;
}
``````

In 2 order to make this work with inputs containing 1 `0`, split the input string into substrings:

``````int main()
{
string num = /* input value*/;

long long greatest_product = 0;
size_t start = -1;
// Iterate over substrings without zero
do {
++start;
size_t end = num.find('0', start);
long long product = greatest(num.substr(start, end-start));
if (greatest_product < product)
greatest_product = product;
start = end;
} while (start != string::npos);

cout << greatest_product << endl;
}
``````
Score: 1
``````public static void main(String[] args) {

String val = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";

int Sum = 0;
String adjacent  = null;
for (int i = 0; i < val.length()-13; i++) {

int total = 1;
for (int j = 0; j < 13; j++) {
total = total * Character.getNumericValue(val.charAt(i+j));
}

if(total > Sum){
Sum = total;
adjacent  = val.substring(i, i+13);
}

}

System.out.println("Sum = " + Sum);

}
``````

0

Score: 1

I had the same problem. int product and 4 int greatestproduct have maximum value it 3 can store since they are 'int' type. They 2 can only store values up to 2147483647.

Use 1 'long long' type instead of 'int'.

Score: 1
``````const thousandDigit =
`73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450`;

const productsOfAdjacentNths = num =>
[...thousandDigit].reduce((acc, _, idx) => {
const nthDigitAdjX = [...thousandDigit.substr(idx, num)].reduce(
(inAcc, inCur) => inCur * inAcc,
1
);

}, []);

``````

0

Score: 1

I solve the problem using this approach

1.Run 7 two loops first outer loop which iterate 6 over numbers in the string.

2.Inner loop 5 reach to next 13 numbers afterwards and 4 store its product in a variable

3.As inner 3 loop ends we only use maximum value.(that 2 what we need) :>

Have a look at my code 1 in C++

``````#include <bits/stdc++.h>
using namespace std;
int main()
{
string s="//thestring is there";
long long unsigned mainans=1,ans=1;           //initialize the value of mainans and ans to 1 (not 0 as product also become 0)
for(int i=0;i<s.length()-13;i++)             //run the loop from 0 upto 13 minus string length
{
ans=1;
for(int j=i;j<i+13;j++)                //now from that particular digit we check 13 digits afterwards it
{
int m=s[j]-'0';                   //convert the number from string to integer
ans*=m;                           //taking product in every interation for 13 digits
}
mainans=max(mainans,ans);            //now taking only max value because that what we need
}
cout<<mainans;
return 0;
}
``````
Score: 0

Although above solutions are good, here 1 is a python implementation which works perfectly:

``````def main():
num=7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450
prod,maxprod=1,0
numtostr=str(num)
length=len(numtostr)
for i in range(1,length-13+1):
prod=1
for z in range(i,i+13):
prod=prod*int(numtostr[z-1:z])
if prod>maxprod:
maxprod=prod
print "Largest 13 digit product is %d" %(maxprod)

if __name__ == '__main__':
main()
``````
Score: 0
``````static long q8(){
long max_product = 1;
long product = 1;
String n = "LONG_INPUT";
for(int i=0;i<13;i++){
max_product *= Integer.parseInt(n.charAt(i)+"");
}
product = max_product;
for(int i=1;i<n.length()-13;i++){
int denom = Integer.parseInt(n.charAt(i-1)+"");
if(denom!=0)
product = product/denom * Integer.parseInt(n.charAt(i+12)+"");
else
product = product(i,n);
max_product = (max_product>product)?max_product:product;
}

return max_product;
}
static long product(int index,String n){
long pro = 1;
for(int i=index;i<index+13;i++){
pro *= Integer.parseInt(n.charAt(i)+"");
}
return pro;
}
``````

0

Score: 0

Try this:

``````    {
DateTime BeganAt = new DateTime();

BeganAt = DateTime.Now;

Int64 result = 0;

string testNumber = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";

StringBuilder StringBuilder = new StringBuilder(13);

Int64 maxNumber = 0;

try
{
char[] numbers = testNumber.ToCharArray();

int tempCounter = 13;

for (int i = 0;i < numbers.Length;i++)
{
if(i < tempCounter)
{
StringBuilder.Append(numbers[i]);
}
else if (i == tempCounter)
{
if (maxNumber < Convert.ToInt64(StringBuilder.ToString()))
{
maxNumber = Convert.ToInt64(StringBuilder.ToString());
}

StringBuilder.Clear();
StringBuilder.Append(numbers[i]);
tempCounter = tempCounter + n;
}
}

result = maxNumber;
}
catch
{
throw;
}

DateTime EndedAt = new DateTime();

EndedAt = DateTime.Now;

TimeSpan TimeItTook = (EndedAt - BeganAt);

return Convert.ToString(result) + " - Time took to execute: " + Convert.ToString(TimeItTook.TotalMilliseconds);
}
``````

0

Score: 0

My functional programming solution

``````def product(number):
product_result = 1
for n in number:
product_result *= int(n)
return product_result

length = 13
indices = range(0, len(number) - 13 + 1)
values = indices
values = map(lambda index: {"index": index, "number": number}, values)
values = map(lambda value: value["number"][value["index"]:value["index"] + length], values)
values = map(product, values)
values = max(values)
print values
``````

0

Score: 0
``````long long int sum = 1,high=0;

for (int i = 0; i < 988; i++)
{
sum = sum*(arr[i] - 48);
for (int j = i + 1; j < i+13; j++)
{
sum = sum*(arr[j]-48);
}
if (sum >= high)
{
high = sum;
}
sum = 1;
}
``````

0

Score: 0

My solution to above problem if someone 6 is still watching this in 2018.Although 5 there are lots of solution to this question 4 here my solution pre-checks the individual 3 13 digits which have a 0 in them.As something 2 multiplied with 0 is always 0 so we can 1 remove this useless computation

``````int j = 0, k = 12;
long long int mult = 1;
long long int max = 0;
char *digits = /*Big number*/
while(k < 1000){
for (int i = j; i <= k; ++i)
{
/* code */
long long int val = digits[i] -'0';
/* check if any number in series contains 0 */
if(val == 0){
break;
}
mult = mult * val;
}
printf("mult is %lld\n",mult );
/* check for max value */
if(max < mult){
max = mult;
}
printf("the new max is %lld\n", max);
j += 1;
k += 1;
mult = 1;
printf("%d iteration finished\n", k);
}
printf("%lld\n", max);
``````
Score: 0

You should use 'int64_t' at the place of 1 'int'

Score: 0

I have solve this problem using python 're' module

Finding 5 all possible 13 digits number without having 4 zeros (total 988 with zero and 283 without 3 zero) then find the product of each of these 2 digits and check for max

here i have used 1 look ahead regex

Note: string must not contain any newline char

s = '731671765...'

``````import re

def product_13(d):
pro = 1
while d:
pro *= d % 10
d //= 10
return pro

pat = re.compile(r'(?=([1-9]{13}))')

all = map(int, re.findall(pat, s))
pro = -1
for i in all:
v = product_13(i)
if pro < v:
pro = v
print(pro)
``````
Score: 0
``````**This is JavaScript solution.**

greatestProductOfAdjacentDigit = (givenNumber, noOfDigit) => {
stringNumberToNumbers = givenNumber => givenNumber.split('').map(each => parseInt(each, 10))

let numbers = mathematicalProblems.stringNumberToNumbers(givenNumber);
return numbers.map(function (each, ind, arr) {
return mathematicalProblems.productOfAll(arr.slice(ind, ind + noOfDigit));
}).reduce(function (accumulator, currentValue) {
return accumulator > currentValue ? accumulator : currentValue;
});
};
``````

Testing

``````const givenNumber = '73167176531330624919225119674426574742355349' +
'194934969835203127745063262395783180169848018694788518438586' +
'156078911294949545950173795833195285320880551112540698747158' +
'523863050715693290963295227443043557668966489504452445231617' +
'318564030987111217223831136222989342338030813533627661428280' +
'644448664523874930358907296290491560440772390713810515859307' +
'960866701724271218839987979087922749219016997208880937766572' +
'733300105336788122023542180975125454059475224352584907711670' +
'556013604839586446706324415722155397536978179778461740649551' +
'492908625693219784686224828397224137565705605749026140797296' +
'865241453510047482166370484403199890008895243450658541227588' +
'666881164271714799244429282308634656748139191231628245861786' +
'645835912456652947654568284891288314260769004224219022671055' +
'626321111109370544217506941658960408071984038509624554443629' +
'812309878799272442849091888458015616609791913387549920052406' +
'368991256071760605886116467109405077541002256983155200055935' +
'72972571636269561882670428252483600823257530420752963450';

``````

0

Score: 0

This is my O(n) approach

``````function findLargest(digits, n){
let largest = 0;
let j = 0;
let res = 1;
for(let i = 0; i< digits.length; i ++){
res = res * parseInt(digits[i]);
if(res == 0){
res = 1;
j = 0;
continue;
}
if(j === n-1){
if(res > largest)
largest = res;
res = res/parseInt(digits[i - j]);
j = j - 1;
}
j = j + 1;
}
return largest;
}
let val = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";

findLargest(val, 13);``````

0

Score: 0

My 2 cents ... Javascript

``````Largest_product_in_series_13 = g => {
return [...g.matchAll(/(?=([1-9]{13}))/g)]
.reduce((a,c)=>(p=eval(c[1].split``.join`*`),p>a?p:a),0)
}
``````

First it matches 2 all 13 digit series, then it reduces to 1 the biggest product.

``````console.log(Largest_product_in_series_13(`7316717653133062491922....
``````

> 23514624000 start: 14.292ms

Score: 0

I solved it using ruby. if

`x = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450"`

then used below 1 logic to solve the problem

``````def print_lps(x)
p = 1
0.upto(x.length-13) do |i|
j = i
xy = 1
while j < i+13 do
xy *= x[j].to_i
j += 1
end
p = xy if xy > p
end
puts p
end

print_lps(x) # 23514624000
``````

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