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  1. Lonely Integer
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Lonely Integer

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  • stephanie_viver2
     + 0 comments
    function lonelyinteger(a: number[]): number {
    if (a.length === 1) return a[0];
    
    const arrMap = new Map<number, number>();
    
    a.forEach((n:number) => (arrMap.has(n)) ? arrMap.set(n, (arrMap.get(n) + 1)) : arrMap.set(n, 1));
    const iterator = arrMap.entries();
    
    for (let i = 0; i < arrMap.size; i++){
        const results = iterator.next().value;
        if (results[1] === 1) return results[0]
    }
}

  • ranechabria
     + 0 comments

    Scala code:

    def lonelyinteger(a: Array[Int]): Int = {
     val distinctElements = a.distinct
     val repeatElements = a.diff(distinctElements)
     val uniqueInteger = (distinctElements diff repeatElements)(0)
     uniqueInteger
    }

  • EhsanAsh
     + 0 comments

    in JavaScript:

    function lonelyinteger(a) {
    const numOfIncidents = a.reduce((obj, int) => {
        obj[int] = (obj[int] || 0) + 1;
        return obj;
    }, {});
    
    for (let key in  numOfIncidents) {
         if (numOfIncidents[key] === 1){
             return parseInt(key);
         }
     }
 }

  • asdert_102345
     + 1 comment

    JAVA: using XOR operator:

    public static int lonelyinteger(List a) {

        int result=a.get(0);
    for(int i=1;i<a.size();i++){
        result^=a.get(i);
    }
    return result;
}

  • hasaladarshana2
     + 0 comments

    in C# public static int lonelyinteger(List a) {

    var cc = a.GroupBy(x => x).ToDictionary(g => g.Key, g => g.ToList());
var dd = cc.Where(x => x.Value.Count == 1).ToList();
return dd[0].Key;

    }