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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• Table of contents
• 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?

Table of contents

---

Task

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?

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

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

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