# Which method returns the smallest integer greater than or equal to a number?

TIL: For certain reasons 🥁 I did a coding challenge on Codility for once and now I want to share my approaches with you.

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• Approach 1: For-Loop
• Approach 2: Map-Function
• Approach 3: Set
• Which method returns the smallest integer greater than or equal to a number?
• Which is the smallest integer greater than the negative integer?
• Which of the following number method returns the smallest integer that is greater than or equal to the arguments in Java?
• What is the smallest positive integer n for which?

Write a function:

``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
7 that, given an array
``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
8 ofN\mathbb{N}N integers, returns the smallest positive integer (greater than 0) that does not occur in A.

For example, given

``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
9, the function should return
``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
0.

Given

``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
1, the function should return 4.

Given

``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
2, the function should return 1.

Write an efficient algorithm for the following assumptions:

N\mathbb{N}N is an integer within the range

``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
3; each element of array A is an integer within the range

``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
0.

## Approach 1: For-Loop

For the first solution we are going to use the for loop.

Step 1: First, we filter the array (which returns a new array) to only get the positive integers, because when only negative integers are in the array, the answer should always return

``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
1.

Step 2: Then we sort the new array in an ascending order.

Step 3: Now, let's create a variable called

``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
2, which stores
``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
1 as a value, because of the reason mentioned before (the smallest possible return is 1).

Step 4: Create the for loop. The for-loop checks if the number in the array is bigger then

``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
2, and when it is, then we already have the solution
``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
1.

Step 5: Otherwise, let's update

``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
2 by the number with which it was compared to increased by
``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
1.

Step 6: When the for-loop is finished, return

``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
2.
``````function solution(A) {
const pos = A.filter(num => num >= 1).sort((a, b) => a - b);
let x = 1;

for(let i = 0; i < pos.length; i++) {
if (x < pos[i]) {
return x;
}
x = pos[i] + 1;
}
return x;
}

console.log(`The solution is \${solution([1, 3, 8, 4, 1, 2])}`);
``````

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## Approach 2: Map-Function

For the second solution we are going to use the map function.

Step 1: We create a variable called

``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
2 like we did in .

Step 2: We use

``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
10 like we did in .

Step 3: Then let's use

``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
11 like we did in .

Step 4: Now we are going to use

``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
12.
``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
12 also creates a new array calling a provided function on every element in the array.

Step 5: Within

``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
12 we again check if
``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
2is smaller then the current number in the array and return it. (Shortcut: If
``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
16 is in the same line the
``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
17, there is no need for
``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
18 and it will return
``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
2.)

Step 6: Otherwise

``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
2 will be updated by the number with which it was compared to increased by
``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
1.

Step 7: When the functionality

``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
2 is returnd.
``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````

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## Approach 3: Set

For the last solution we are going to use

``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
73 method.

Step 1: Create a variable called

``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
74, and store a new instance of
``````function solution(A) {
let x = 1

A.filter(x => x >= 1)
.sort((a, b) => a - b)
.map((val, i, arr) => {
if(x < arr[i]) return
x = arr[i] + 1
})
return x
}

console.log(`The solution is \${solution2([-1, 3, 8, 6, 1, 2])}`);
``````
75 with the array.

### What method is used to find the smallest integer value that is greater than or equal to the argument or mathematical integer?

ceil () is used to find the smallest integer value that is greater than or equal to the argument or mathematical integer.

### What method returns the smallest integer greater than or equal to a decimal number?

Ceiling(Decimal) Returns the smallest integral value that is greater than or equal to the specified decimal number.

### Which of the following number method returns the smallest integer that is greater than or equal to the arguments in Java?

ceil() method. This method returns the smallest integer greater than or equal to a given number.

### Which method returns the next integer greater than or equal to a given number?

ceil() The Math. ceil() function always rounds up and returns the smaller integer greater than or equal to a given number.