Linear Search

Easy Searching & Sorting

Search an array for a target value and return its index (or -1 if not found).

Sample input

[4, 2, 7, 1, 9], target = 7

Sample output

Solution

arr = [4, 2, 7, 1, 9]
target = 7
index = -1
for i, x in enumerate(arr):
    if x == target:
        index = i
        break
print(index)

const arr = [4, 2, 7, 1, 9];
const target = 7;
let index = -1;
for (let i = 0; i < arr.length; i++) {
  if (arr[i] === target) {
    index = i;
    break;
  }
}
console.log(index);

public class Main {
    public static void main(String[] args) {
        int[] arr = {4, 2, 7, 1, 9};
        int target = 7, index = -1;
        for (int i = 0; i < arr.length; i++) {
            if (arr[i] == target) {
                index = i;
                break;
            }
        }
        System.out.println(index);
    }
}

fun main() {
    val arr = intArrayOf(4, 2, 7, 1, 9)
    val target = 7
    var index = -1
    for (i in arr.indices) {
        if (arr[i] == target) {
            index = i
            break
        }
    }
    println(index)
}

let arr = [4, 2, 7, 1, 9]
let target = 7
var index = -1
for i in 0..<arr.count where arr[i] == target {
    index = i
    break
}
print(index)

void main() {
  List<int> arr = [4, 2, 7, 1, 9];
  int target = 7;
  int index = -1;
  for (int i = 0; i < arr.length; i++) {
    if (arr[i] == target) {
      index = i;
      break;
    }
  }
  print(index);
}

#include <iostream>
using namespace std;

int main() {
    int arr[] = {4, 2, 7, 1, 9};
    int n = 5, target = 7, index = -1;
    for (int i = 0; i < n; i++) {
        if (arr[i] == target) {
            index = i;
            break;
        }
    }
    cout << index << endl;
    return 0;
}

#include <stdio.h>

int main() {
    int arr[] = {4, 2, 7, 1, 9};
    int n = 5, target = 7, index = -1;
    for (int i = 0; i < n; i++) {
        if (arr[i] == target) {
            index = i;
            break;
        }
    }
    printf("%d\n", index);
    return 0;
}