Filtering an array based on two specified boundary conditions

Question

Filtering an array based on two specified boundary conditions

I have an array containing numbers from 1 to 7

const num = [1,2,3,4,5,6,7];

and two boundary values:

var first = 6
var second = 3

The desired output should be:

[6,7,1,2,3]

Explanation:

The resulting array should combine two sub-arrays:

Numbers greater than or equal to 'first' (6) on the left side: [6, 7]
Numbers less than or equal to 'second' (3) on the right side: [1, 2, 3]

I attempted to achieve this with the following code but the positions of elements are not as expected:

const result = num.filter((n,p) => {
  if(p >= num.indexOf(first)) {return n}
  else if(p <= num.indexOf(second)) {return n}
});

console.log(result) // [1,2,3,6,7]

An alternative method involves looping through the array twice, which is not efficient:

const num = [1,2,3,4,5,6,7];
var first = 6
var second = 3
var arr = []
num.forEach((n,p) => {
  if(p >= num.indexOf(first)) {arr.push(n)}
});
num.forEach((n,p) => {
  if(p <= num.indexOf(second)) {arr.push(n)}
});

console.log(arr) //[6,7,1,2,3]

Is there a better way to achieve this?

javascript arrays loops foreach

Answer 1

Answer №1

var numbers = [1, 2, 3, 4, 5, 6, 7];
var firstNum = 6;
var secondNum = 3;
const result = [
  ...numbers.filter((num) => num >= firstNum),
  ...numbers.filter((num) => num <= secondNum)
];
console.log(result); // [6,7,1,2,3]

Modified Version:

Iterate through the array once:

var numbers = [1, 2, 3, 4, 5, 6, 7];
var firstNum = 6;
var secondNum = 3;
var leftArray = [];
var rightArray = [];
numbers.forEach((num) => {
  if (num >= firstNum) leftArray.push(num);
  if (num <= secondNum) rightArray.push(num);
});
const finalResult = [...leftArray, ...rightArray];
console.log(finalResult); // [6,7,1,2,3]

Answer 2

var numbers = [1, 2, 3, 4, 5, 6, 7];
var firstNum = 6;
var secondNum = 3;
const result = [
  ...numbers.filter((num) => num >= firstNum),
  ...numbers.filter((num) => num <= secondNum)
];
console.log(result); // [6,7,1,2,3]

Modified Version:

Iterate through the array once:

var numbers = [1, 2, 3, 4, 5, 6, 7];
var firstNum = 6;
var secondNum = 3;
var leftArray = [];
var rightArray = [];
numbers.forEach((num) => {
  if (num >= firstNum) leftArray.push(num);
  if (num <= secondNum) rightArray.push(num);
});
const finalResult = [...leftArray, ...rightArray];
console.log(finalResult); // [6,7,1,2,3]

Answer 3

Answer №2

Your inquiry appears to be quite unique, but the solution provided below should meet your needs:

const arr=[1,2,3,4,5,6,7];

// looking at positions of elements:
const getSubArraysPos=(ar,m,n)=>[...ar,...ar].slice(m-1,ar.length+n);
console.log(getSubArraysPos(arr,6,3))

// looking at array values:
const getSubArraysVal=(ar,m,n)=>[].concat.call(...ar.reduce((a,c)=>{
  if (m>n) { if(c>=m) a[0].push(c); else if (c<=n) a[1].push(c); }
  else if(c>=m && c<=n) a[0].push(c);
  return a }, [[],[]]));
  
console.log("6,3:",getSubArraysPos(arr,6,3));
console.log("5,2:",getSubArraysVal(arr,5,2));
console.log("2,5:",getSubArraysVal(arr,2,5))

The code may seem a bit concise, but the goal was to condense it into as few lines as possible.

Answer 4

Your inquiry appears to be quite unique, but the solution provided below should meet your needs:

const arr=[1,2,3,4,5,6,7];

// looking at positions of elements:
const getSubArraysPos=(ar,m,n)=>[...ar,...ar].slice(m-1,ar.length+n);
console.log(getSubArraysPos(arr,6,3))

// looking at array values:
const getSubArraysVal=(ar,m,n)=>[].concat.call(...ar.reduce((a,c)=>{
  if (m>n) { if(c>=m) a[0].push(c); else if (c<=n) a[1].push(c); }
  else if(c>=m && c<=n) a[0].push(c);
  return a }, [[],[]]));
  
console.log("6,3:",getSubArraysPos(arr,6,3));
console.log("5,2:",getSubArraysVal(arr,5,2));
console.log("2,5:",getSubArraysVal(arr,2,5))

The code may seem a bit concise, but the goal was to condense it into as few lines as possible.

Answer 5

Answer №3

To begin with, let me clarify the condition: the desired outcome entails merging 2 arrays:

>= first on the left [6, 7]
<= second on the right [1,2,3]

Result:1 & 2 = [6,7,1,2,3]

Approach:

An efficient method involves utilizing Array#reduce by initializing a result object as follows: {first:[], second: []}. Then, employ spread to concatenate the two arrays.

This approach ensures a time complexity of O(n).

const num = [1,2,3,4,5,6,7];
var first = 6, second = 3;

const result = num.reduce((acc, curr) => {
  if(curr >= first) acc.first.push(curr);
  if(curr <= second) acc.second.push(curr);
  
  return acc;
}, {first:[], second: []});
console.log([...result.first, ...result.second]);

Answer 6

To begin with, let me clarify the condition: the desired outcome entails merging 2 arrays:

>= first on the left [6, 7]
<= second on the right [1,2,3]

Result:1 & 2 = [6,7,1,2,3]

Approach:

An efficient method involves utilizing Array#reduce by initializing a result object as follows: {first:[], second: []}. Then, employ spread to concatenate the two arrays.

This approach ensures a time complexity of O(n).

const num = [1,2,3,4,5,6,7];
var first = 6, second = 3;

const result = num.reduce((acc, curr) => {
  if(curr >= first) acc.first.push(curr);
  if(curr <= second) acc.second.push(curr);
  
  return acc;
}, {first:[], second: []});
console.log([...result.first, ...result.second]);

Answer 7

Answer №4

While your initial attempt using the .filter method was somewhat successful, it's important to note that .filter doesn't have the ability to rearrange elements within an array.

To solve this issue, you can identify the index of the first specified value and use it as a reference point to extract part of the array. Then, locate the index of the second value, slice off the excess portion at the beginning of the array, and finally combine the two sections together.

const secondIdx = num.indexOf(second);
const firstIdx = num.indexOf(first, second+1);
const result = num.slice(firstIdx).concat(num.slice(0, secondIdx+1));

Answer 8

While your initial attempt using the .filter method was somewhat successful, it's important to note that .filter doesn't have the ability to rearrange elements within an array.

To solve this issue, you can identify the index of the first specified value and use it as a reference point to extract part of the array. Then, locate the index of the second value, slice off the excess portion at the beginning of the array, and finally combine the two sections together.

const secondIdx = num.indexOf(second);
const firstIdx = num.indexOf(first, second+1);
const result = num.slice(firstIdx).concat(num.slice(0, secondIdx+1));

Answer 9

Answer №5

I feel the need to provide my own answer because the previous answers seem to have some sort of issue.

function arr(num, first, second){
 let arr = num.reduce((acc, v, i) => {
  let i1 = num.indexOf(first)
  let i2 = num.indexOf(second)
  if(i1 < i2 && i >= i1 && i <= i2) acc['a'].push(v)
  if(i1 > i2 && i <= i2){
    acc['t1'].push(v) 
  } 
  if(i1 > i2 && i >= i1) {
    acc['t2'].push(v) 
  }
  if(i1 > i2) acc['a'] = [...acc['t2'], ...acc['t1']]
  return acc
}, {a : [], t1: [], t2: []})

return arr['a']
}

let num = [1,2,3,4,5,6,7];
console.log(arr(num,2,5))
console.log(arr(num,5,2))
console.log(arr(num,3,6))
console.log(arr(num,6,3))

Answer 10

I feel the need to provide my own answer because the previous answers seem to have some sort of issue.

function arr(num, first, second){
 let arr = num.reduce((acc, v, i) => {
  let i1 = num.indexOf(first)
  let i2 = num.indexOf(second)
  if(i1 < i2 && i >= i1 && i <= i2) acc['a'].push(v)
  if(i1 > i2 && i <= i2){
    acc['t1'].push(v) 
  } 
  if(i1 > i2 && i >= i1) {
    acc['t2'].push(v) 
  }
  if(i1 > i2) acc['a'] = [...acc['t2'], ...acc['t1']]
  return acc
}, {a : [], t1: [], t2: []})

return arr['a']
}

let num = [1,2,3,4,5,6,7];
console.log(arr(num,2,5))
console.log(arr(num,5,2))
console.log(arr(num,3,6))
console.log(arr(num,6,3))

Filtering an array based on two specified boundary conditions

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

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