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let arr = [1, 2, 3, 4]
for (let i = 0; i < arr.length; i++) {
console.log(arr[i])
}
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const numbers = [1, 2, 3, 4, 5];
for (i = 0; i < numbers.length; i++) {
console.log(numbers[i]);
}
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/* new options with IE6: loop through array of objects */
const people = [
{id: 100, name: 'Vikash'},
{id: 101, name: 'Sugam'},
{id: 102, name: 'Ashish'}
];
// using for of
for (let persone of people) {
console.log(persone.id + ': ' + persone.name);
}
// using forEach(...)
people.forEach(person => {
console.log(persone.id + ': ' + persone.name);
});
// output of above two methods
// 100: Vikash
// 101: Sugam
// 102: Ashish
// forEach(...) with index
people.forEach((person, index) => {
console.log(index + ': ' + persone.name);
});
// output of above code in console
// 0: Vikash
// 1: Sugam
// 2: Ashish
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var arr = [1, 2, 3, 4, 5];
for (var i = arr.length - 1; i >= 0; i--) {
console.log(arr[i]);
}
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var min = arr[0];
var max = arr[0];
for(var i=1; i<arr.length; i++){
if(arr[i] < min){
min = arr[i];
}
if(arr[i] > max){
max = arr[i];
}
return [min, max];
}
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var arr = ['a', 'b', 'c'];
arr.forEach(item => {
console.log(item);
});
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You may assume that the sequence is always correct, i.e., every booked room was previously free, and every freed room was previously booked.
In case, 2 rooms have been booked the same number of times, you have to return Lexographically smaller room.
A string 'a' is lexicographically smaller than a string 'b' (of the same length) if in the first position where 'a' and 'b' differ, string 'a' has a letter that appears earlier in the alphabet than the corresponding letter in string 'b'. For example, "abcd" is lexicographically smaller than "acbd" because the first position they differ in is at the second letter, and 'b' comes before 'c'.
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var temp = [];
for(var i=0; i<arr1.length; i++){
temp.push(arr1[i]);
}
for(var i=0; i<arr2.length; i++){
temp.push(arr2[i]);
}
console.log('temp is now', temp);
return temp;
}