Duplicated and missing number in an array

-3

The input nums is supposed to be an array of unique integers ranging from 1 to nums.length (inclusive). However, there is a mistake: one of the numbers in the array duplicated, which means another number is missing. Find and return the sum of the duplicated number and the missing number. Example in the array [4, 3, 3, 1], 3is present twice and 2 is missing, so 3 +2= 5 should be returned. // please write return statement//no matter which language its up to you

4/26/2021 2:10:57 AM

Pavan priya

7 Answers

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let duplicateValue = 0 let missingValue = 0 let sortData = arr.sort((a,b)=>b>a?-1:1) sortData.forEach((ele,index)=>{ if(arr.indexOf(ele) !== index){ duplicateValue = ele } if(ele !== index+1){ missingValue = index + 1 } }) console.log(duplicateValue+missingValue)

Gautam

At least specify a relevant language in the tags above rather than a copy of the question. https://code.sololearn.com/W3uiji9X28C1/?ref=app

Ipang

Here's an algorithmic approach example that should be easy to rewrite in any language, with an explanation for how to achieve O(n) Time and O(1) space complexities. https://code.sololearn.com/cYmujdEYEsS9/?ref=app

ChaoticDawg

In Which language you wrote this code

hima bindu

let arr=[1,2,3,5,5,6] for(let i=0;i<arr.length;i++){ if(arr[i+1]==arr[i]){ console.log(arr[i]+(i+1)) } }

Gautam

Hii

Pompi Nath

-1

l=[1,3,3,4] c=set([x for x in l if l.count(x)>1]) rep_nums=[] for i in c: rep_nums.append(i) missing_num = 0 for i in range(len(l)): if i+1 == l[i]: pass else: missing_num = i+1 print(missing_num+sum(rep_nums)) HOPE IT WORKS!!!

sarada lakshmi