Question

How to find a number in array that appears once in JavaScript

Answer 1

function number_exist_few_times(arr, number, index) {
    for (let i = 0; i < arr.length; i++) {
        if (arr[i] == number && i != index) {
            return true;
        }
    }
    
    return false;
}
   
function get_single_number(arr) { 
     for (let i = 0; i < arr.length; i++) {
        if (!number_exist_few_times(arr, arr[i], i)) {
            return arr[i];
        }
    }
    
    return -1;
} 
    
 
const arr = new Array(3, 2, 1, 2, 3, 3, 1, 5, 9, 7, 7, 9, 9); 
       
console.log(get_single_number(arr)); 
   
   
   
/*
run:
   
5
   
*/

 

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Answer 2

// indexOf(searchElement, fromIndex)

function get_single_number(arr) {
    for (let i = 0; i < arr.length; i++) {
        if (arr.indexOf(arr[i], arr.indexOf(arr[i]) + 1) === -1) {
            return arr[i];
        }    
    };
    
    return -1;
};
    
 
const arr = new Array(3, 2, 1, 2, 3, 3, 1, 5, 9, 7, 7, 9, 9); 
       
console.log(get_single_number(arr)); 

   
   
/*
run:
   
5
   
*/

 

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