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javascript reduce(),keys(), filter() method

While solving a problem to find missing numbers from an array.

let arr = [15,3,5,6,7,2,1,8,12];

let findMissing = (arr)=>{
    let res = arr.reduce((acc,n)=>{
        acc[n]=1;
        return acc;
    },[])
    
    return [...res.keys()].filter((i)=>{
        if(res[i]!==1) return i
    })
}

console.log(findMissing(arr));

output:[4, 9, 10, 11, 13, 14]

Issue is why output does not have zero in it: [0,4, 9, 10, 11, 13, 14]

Tags:

2

Answers


  1. 0

    because you return i in the filter method (you return 0).
    i === 0 && 0 == false it is filtered out because when returning a falsy value for a filter function it will strip it.

    if you return true like in my example it works.

    let arr = [15,3,5,6,7,2,1,8,12];

let findMissing = (arr)=>{
    let res = arr.reduce((acc,n)=>{
        acc[n]=1;
        return acc;
    },[])
    
    return [...res.keys()].filter((i)=>{
        return res[i] !== 1
    })
}

console.log(findMissing(arr));
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  2. 0

    Alternatively, findMissing() can be written as a one-liner, using .reduce() and .includes():

    let arr = [15,3,5,6,7,2,1,8,12];

let findMissing=arr=>
  Array(Math.max(...arr)+1).fill(0)
  .reduce((a,_,i)=>(!arr.includes(i)&&a.push(i),a),[]);

console.log(findMissing(arr));

    However, the performance could be improved by creating a Set s from arr with s=new Set(arr) and then using !s.has(i) instead of !arr.includes(i) inside the reduce() callback function.

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