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Check Digit Sum Javascript- recursion
Adding digits from a number, using recursivity - javascriptHow do JavaScript closures work?How do I check if an element is hidden in jQuery?How do I remove a property from a JavaScript object?How do I check if an array includes an object in JavaScript?Which equals operator (== vs ===) should be used in JavaScript comparisons?Setting “checked” for a checkbox with jQuery?How do I include a JavaScript file in another JavaScript file?What does “use strict” do in JavaScript, and what is the reasoning behind it?How to check whether a string contains a substring in JavaScript?How do I remove a particular element from an array in JavaScript?
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Looking for Javascript solution in recursion to get the sum of all digits in number until single digit come as result
For example, for the number is "55555" the sum of all digits is 25. Because this is not a single-digit number, 2 and 5 would be added, and the result, 7.
I tried the below solution based on the algorithm.
function getSum(n) sum > 9)
if(n == 0)
n = sum;
sum = 0;
sum += n % 10;
n /= 10;
return sum;
console.log(getSum("55555"));
javascript
asked Nov 13 '18 at 20:33
A G JA G J
21
add a comment |
Looking for Javascript solution in recursion to get the sum of all digits in number until single digit come as result
For example, for the number is "55555" the sum of all digits is 25. Because this is not a single-digit number, 2 and 5 would be added, and the result, 7.
I tried the below solution based on the algorithm.
function getSum(n) sum > 9)
if(n == 0)
n = sum;
sum = 0;
sum += n % 10;
n /= 10;
return sum;
console.log(getSum("55555"));
javascript
asked Nov 13 '18 at 20:33
A G JA G J
21
Check here I have found same problem here
– Dulanga Heshan
Nov 13 '18 at 20:55
add a comment |
Looking for Javascript solution in recursion to get the sum of all digits in number until single digit come as result
For example, for the number is "55555" the sum of all digits is 25. Because this is not a single-digit number, 2 and 5 would be added, and the result, 7.
I tried the below solution based on the algorithm.
function getSum(n) sum > 9)
if(n == 0)
n = sum;
sum = 0;
sum += n % 10;
n /= 10;
return sum;
console.log(getSum("55555"));
javascript
asked Nov 13 '18 at 20:33
A G JA G J
21
Looking for Javascript solution in recursion to get the sum of all digits in number until single digit come as result
For example, for the number is "55555" the sum of all digits is 25. Because this is not a single-digit number, 2 and 5 would be added, and the result, 7.
I tried the below solution based on the algorithm.
function getSum(n) sum > 9)
if(n == 0)
n = sum;
sum = 0;
sum += n % 10;
n /= 10;
return sum;
console.log(getSum("55555"));
javascript
javascript
asked Nov 13 '18 at 20:33
A G JA G J
21
asked Nov 13 '18 at 20:33
A G JA G J
21
edited Nov 13 '18 at 20:44
A G J
asked Nov 13 '18 at 20:33
A G JA G J
21
asked Nov 13 '18 at 20:33
A G JA G J
21
asked Nov 13 '18 at 20:33
A G JA G J
21
21
Check here I have found same problem here
– Dulanga Heshan
Nov 13 '18 at 20:55
add a comment |
Check here I have found same problem here
– Dulanga Heshan
Nov 13 '18 at 20:55
Check here I have found same problem here
– Dulanga Heshan
Nov 13 '18 at 20:55
Check here I have found same problem here
– Dulanga Heshan
Nov 13 '18 at 20:55
add a comment |
3 Answers
3
active
oldest
votes
This would kind of work, but I'm almost sure there's a beautiful one-line solution which I just don't see yet.
function singleDigitSum(str)
str = [...str].reduce((acc, c) => return Number(c) + acc , 0)
while (str.toString().length > 1)
str = singleDigitSum(str.toString());
return str
console.log(singleDigitSum("55555"))Explanation:
As a first step in your function you reassign to the parameter passed to the function the result of a reducer function which sums up all numbers in your String. To be able to use Array.prototype.reduce() function, I'm spreading your str into an array using [...str].
Then, for as often as that reducer returns a value with more than one digit, rinse and repeat. When the while loop exits, the result is single digit and can be returned.
answered Nov 13 '18 at 20:48
connexoconnexo
23.6k93863
it's not a pure recursion, you are using one another functionreducehere .
– Jaisa Ram
Nov 13 '18 at 21:57
add a comment |
function checSumOfDigit(num, sum = "0")
if (num.length == 1 && sum.length !== 1)
return checSumOfDigit(Number(sum) + Number(num) + "", "0");
else if (num.length == 1)
return Number(sum) + Number(num);
num = num.split("")
sum = Number(sum) + Number(num.pop());
return checSumOfDigit(num.join(""), sum + "")
console.log(checSumOfDigit("567"));
console.log(checSumOfDigit("123"));
console.log(checSumOfDigit("55555"));this code might be help you
answered Nov 13 '18 at 22:13
Jaisa RamJaisa Ram
27438
add a comment |
If you need a recursion try this one
function CheckDigitSum(number)
let nums = number.split('');
if (nums.length > 1)
let sum = 0;
for (let i = 0; i < nums.length; i++)
sum += Number(nums[i]);
return CheckDigitSum(sum.toString());
else
return parseInt(nums[0], 10);
answered Mar 23 at 4:36
Daniel DeulinDaniel Deulin
316
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
This would kind of work, but I'm almost sure there's a beautiful one-line solution which I just don't see yet.
function singleDigitSum(str)
str = [...str].reduce((acc, c) => return Number(c) + acc , 0)
while (str.toString().length > 1)
str = singleDigitSum(str.toString());
return str
console.log(singleDigitSum("55555"))Explanation:
As a first step in your function you reassign to the parameter passed to the function the result of a reducer function which sums up all numbers in your String. To be able to use Array.prototype.reduce() function, I'm spreading your str into an array using [...str].
Then, for as often as that reducer returns a value with more than one digit, rinse and repeat. When the while loop exits, the result is single digit and can be returned.
answered Nov 13 '18 at 20:48
connexoconnexo
23.6k93863
it's not a pure recursion, you are using one another functionreducehere .
– Jaisa Ram
Nov 13 '18 at 21:57
add a comment |
This would kind of work, but I'm almost sure there's a beautiful one-line solution which I just don't see yet.
function singleDigitSum(str)
str = [...str].reduce((acc, c) => return Number(c) + acc , 0)
while (str.toString().length > 1)
str = singleDigitSum(str.toString());
return str
console.log(singleDigitSum("55555"))Explanation:
As a first step in your function you reassign to the parameter passed to the function the result of a reducer function which sums up all numbers in your String. To be able to use Array.prototype.reduce() function, I'm spreading your str into an array using [...str].
Then, for as often as that reducer returns a value with more than one digit, rinse and repeat. When the while loop exits, the result is single digit and can be returned.
answered Nov 13 '18 at 20:48
connexoconnexo
23.6k93863
it's not a pure recursion, you are using one another functionreducehere .
– Jaisa Ram
Nov 13 '18 at 21:57
add a comment |
This would kind of work, but I'm almost sure there's a beautiful one-line solution which I just don't see yet.
function singleDigitSum(str)
str = [...str].reduce((acc, c) => return Number(c) + acc , 0)
while (str.toString().length > 1)
str = singleDigitSum(str.toString());
return str
console.log(singleDigitSum("55555"))Explanation:
As a first step in your function you reassign to the parameter passed to the function the result of a reducer function which sums up all numbers in your String. To be able to use Array.prototype.reduce() function, I'm spreading your str into an array using [...str].
Then, for as often as that reducer returns a value with more than one digit, rinse and repeat. When the while loop exits, the result is single digit and can be returned.
answered Nov 13 '18 at 20:48
connexoconnexo
23.6k93863
This would kind of work, but I'm almost sure there's a beautiful one-line solution which I just don't see yet.
function singleDigitSum(str)
str = [...str].reduce((acc, c) => return Number(c) + acc , 0)
while (str.toString().length > 1)
str = singleDigitSum(str.toString());
return str
console.log(singleDigitSum("55555"))Explanation:
As a first step in your function you reassign to the parameter passed to the function the result of a reducer function which sums up all numbers in your String. To be able to use Array.prototype.reduce() function, I'm spreading your str into an array using [...str].
Then, for as often as that reducer returns a value with more than one digit, rinse and repeat. When the while loop exits, the result is single digit and can be returned.
function singleDigitSum(str)
str = [...str].reduce((acc, c) => return Number(c) + acc , 0)
while (str.toString().length > 1)
str = singleDigitSum(str.toString());
return str
console.log(singleDigitSum("55555"))function singleDigitSum(str)
str = [...str].reduce((acc, c) => return Number(c) + acc , 0)
while (str.toString().length > 1)
str = singleDigitSum(str.toString());
return str
console.log(singleDigitSum("55555"))answered Nov 13 '18 at 20:48
connexoconnexo
23.6k93863
edited Nov 13 '18 at 20:53
answered Nov 13 '18 at 20:48
connexoconnexo
23.6k93863
answered Nov 13 '18 at 20:48
connexoconnexo
23.6k93863
answered Nov 13 '18 at 20:48
connexoconnexo
23.6k93863
23.6k93863
it's not a pure recursion, you are using one another functionreducehere .
– Jaisa Ram
Nov 13 '18 at 21:57
add a comment |
it's not a pure recursion, you are using one another functionreducehere .
– Jaisa Ram
Nov 13 '18 at 21:57
it's not a pure recursion, you are using one another function
reduce here .– Jaisa Ram
Nov 13 '18 at 21:57
it's not a pure recursion, you are using one another function
reduce here .– Jaisa Ram
Nov 13 '18 at 21:57
add a comment |
function checSumOfDigit(num, sum = "0")
if (num.length == 1 && sum.length !== 1)
return checSumOfDigit(Number(sum) + Number(num) + "", "0");
else if (num.length == 1)
return Number(sum) + Number(num);
num = num.split("")
sum = Number(sum) + Number(num.pop());
return checSumOfDigit(num.join(""), sum + "")
console.log(checSumOfDigit("567"));
console.log(checSumOfDigit("123"));
console.log(checSumOfDigit("55555"));this code might be help you
answered Nov 13 '18 at 22:13
Jaisa RamJaisa Ram
27438
add a comment |
function checSumOfDigit(num, sum = "0")
if (num.length == 1 && sum.length !== 1)
return checSumOfDigit(Number(sum) + Number(num) + "", "0");
else if (num.length == 1)
return Number(sum) + Number(num);
num = num.split("")
sum = Number(sum) + Number(num.pop());
return checSumOfDigit(num.join(""), sum + "")
console.log(checSumOfDigit("567"));
console.log(checSumOfDigit("123"));
console.log(checSumOfDigit("55555"));this code might be help you
answered Nov 13 '18 at 22:13
Jaisa RamJaisa Ram
27438
add a comment |
function checSumOfDigit(num, sum = "0")
if (num.length == 1 && sum.length !== 1)
return checSumOfDigit(Number(sum) + Number(num) + "", "0");
else if (num.length == 1)
return Number(sum) + Number(num);
num = num.split("")
sum = Number(sum) + Number(num.pop());
return checSumOfDigit(num.join(""), sum + "")
console.log(checSumOfDigit("567"));
console.log(checSumOfDigit("123"));
console.log(checSumOfDigit("55555"));this code might be help you
answered Nov 13 '18 at 22:13
Jaisa RamJaisa Ram
27438
function checSumOfDigit(num, sum = "0")
if (num.length == 1 && sum.length !== 1)
return checSumOfDigit(Number(sum) + Number(num) + "", "0");
else if (num.length == 1)
return Number(sum) + Number(num);
num = num.split("")
sum = Number(sum) + Number(num.pop());
return checSumOfDigit(num.join(""), sum + "")
console.log(checSumOfDigit("567"));
console.log(checSumOfDigit("123"));
console.log(checSumOfDigit("55555"));this code might be help you
function checSumOfDigit(num, sum = "0")
if (num.length == 1 && sum.length !== 1)
return checSumOfDigit(Number(sum) + Number(num) + "", "0");
else if (num.length == 1)
return Number(sum) + Number(num);
num = num.split("")
sum = Number(sum) + Number(num.pop());
return checSumOfDigit(num.join(""), sum + "")
console.log(checSumOfDigit("567"));
console.log(checSumOfDigit("123"));
console.log(checSumOfDigit("55555"));function checSumOfDigit(num, sum = "0")
if (num.length == 1 && sum.length !== 1)
return checSumOfDigit(Number(sum) + Number(num) + "", "0");
else if (num.length == 1)
return Number(sum) + Number(num);
num = num.split("")
sum = Number(sum) + Number(num.pop());
return checSumOfDigit(num.join(""), sum + "")
console.log(checSumOfDigit("567"));
console.log(checSumOfDigit("123"));
console.log(checSumOfDigit("55555"));answered Nov 13 '18 at 22:13
Jaisa RamJaisa Ram
27438
answered Nov 13 '18 at 22:13
Jaisa RamJaisa Ram
27438
answered Nov 13 '18 at 22:13
Jaisa RamJaisa Ram
27438
answered Nov 13 '18 at 22:13
Jaisa RamJaisa Ram
27438
27438
add a comment |
add a comment |
If you need a recursion try this one
function CheckDigitSum(number)
let nums = number.split('');
if (nums.length > 1)
let sum = 0;
for (let i = 0; i < nums.length; i++)
sum += Number(nums[i]);
return CheckDigitSum(sum.toString());
else
return parseInt(nums[0], 10);
answered Mar 23 at 4:36
Daniel DeulinDaniel Deulin
316
add a comment |
If you need a recursion try this one
function CheckDigitSum(number)
let nums = number.split('');
if (nums.length > 1)
let sum = 0;
for (let i = 0; i < nums.length; i++)
sum += Number(nums[i]);
return CheckDigitSum(sum.toString());
else
return parseInt(nums[0], 10);
answered Mar 23 at 4:36
Daniel DeulinDaniel Deulin
316
add a comment |
If you need a recursion try this one
function CheckDigitSum(number)
let nums = number.split('');
if (nums.length > 1)
let sum = 0;
for (let i = 0; i < nums.length; i++)
sum += Number(nums[i]);
return CheckDigitSum(sum.toString());
else
return parseInt(nums[0], 10);
answered Mar 23 at 4:36
Daniel DeulinDaniel Deulin
316
If you need a recursion try this one
function CheckDigitSum(number)
let nums = number.split('');
if (nums.length > 1)
let sum = 0;
for (let i = 0; i < nums.length; i++)
sum += Number(nums[i]);
return CheckDigitSum(sum.toString());
else
return parseInt(nums[0], 10);
answered Mar 23 at 4:36
Daniel DeulinDaniel Deulin
316
edited Mar 24 at 3:36
answered Mar 23 at 4:36
Daniel DeulinDaniel Deulin
316
answered Mar 23 at 4:36
Daniel DeulinDaniel Deulin
316
answered Mar 23 at 4:36
Daniel DeulinDaniel Deulin
316
316
add a comment |
add a comment |
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Check here I have found same problem here
– Dulanga Heshan
Nov 13 '18 at 20:55