how to make array of objects from array based on some condtion in javascript

-3

i have an array like let x = [1,5,2,6,7,9]

but i want to make this as array of objects like below code map or JSON

   let y = [

             { st:1,ed:2}, 

             {st:5,ed:7},

             {st:9,ed:9}

             ]

based on the continuity of the digits, any help or suggestion required

asked Nov 24 '18 at 12:00

Hemanth SP

447

2

what happens to 6? why the random order?

– Nina Scholz
Nov 24 '18 at 12:02

@Nina Scholz no need of 6 it just {st:5,ed:7} interview question, needed only start and end

– Hemanth SP
Nov 24 '18 at 12:03

This makes absolutely no sense to me ... unless I'm missing some piece of the puzzle.

– techouse
Nov 24 '18 at 12:03

1

what does it mean based on the continuity of the digits?

– Dennis Vash
Nov 24 '18 at 12:12

add a comment |

-3

i have an array like let x = [1,5,2,6,7,9]

but i want to make this as array of objects like below code map or JSON

   let y = [

             { st:1,ed:2}, 

             {st:5,ed:7},

             {st:9,ed:9}

             ]

based on the continuity of the digits, any help or suggestion required

asked Nov 24 '18 at 12:00

Hemanth SP

447

2

what happens to 6? why the random order?

– Nina Scholz
Nov 24 '18 at 12:02

@Nina Scholz no need of 6 it just {st:5,ed:7} interview question, needed only start and end

– Hemanth SP
Nov 24 '18 at 12:03

This makes absolutely no sense to me ... unless I'm missing some piece of the puzzle.

– techouse
Nov 24 '18 at 12:03

1

what does it mean based on the continuity of the digits?

– Dennis Vash
Nov 24 '18 at 12:12

add a comment |

-3

i have an array like let x = [1,5,2,6,7,9]

but i want to make this as array of objects like below code map or JSON

   let y = [

             { st:1,ed:2}, 

             {st:5,ed:7},

             {st:9,ed:9}

             ]

based on the continuity of the digits, any help or suggestion required

asked Nov 24 '18 at 12:00

Hemanth SP

447

i have an array like let x = [1,5,2,6,7,9]

but i want to make this as array of objects like below code map or JSON

   let y = [

             { st:1,ed:2}, 

             {st:5,ed:7},

             {st:9,ed:9}

             ]

based on the continuity of the digits, any help or suggestion required

javascript

asked Nov 24 '18 at 12:00

Hemanth SP

447

asked Nov 24 '18 at 12:00

Hemanth SP

447

asked Nov 24 '18 at 12:00

Hemanth SP

447

asked Nov 24 '18 at 12:00

Hemanth SP

447

asked Nov 24 '18 at 12:00

Hemanth SP

447

2

what happens to 6? why the random order?

– Nina Scholz
Nov 24 '18 at 12:02

@Nina Scholz no need of 6 it just {st:5,ed:7} interview question, needed only start and end

– Hemanth SP
Nov 24 '18 at 12:03

This makes absolutely no sense to me ... unless I'm missing some piece of the puzzle.

– techouse
Nov 24 '18 at 12:03

1

what does it mean based on the continuity of the digits?

– Dennis Vash
Nov 24 '18 at 12:12

add a comment |

2

what happens to 6? why the random order?

– Nina Scholz
Nov 24 '18 at 12:02

@Nina Scholz no need of 6 it just {st:5,ed:7} interview question, needed only start and end

– Hemanth SP
Nov 24 '18 at 12:03

This makes absolutely no sense to me ... unless I'm missing some piece of the puzzle.

– techouse
Nov 24 '18 at 12:03

1

what does it mean based on the continuity of the digits?

– Dennis Vash
Nov 24 '18 at 12:12

what happens to 6? why the random order?

– Nina Scholz
Nov 24 '18 at 12:02

@Nina Scholz no need of 6 it just {st:5,ed:7} interview question, needed only start and end

– Hemanth SP
Nov 24 '18 at 12:03

This makes absolutely no sense to me ... unless I'm missing some piece of the puzzle.

– techouse
Nov 24 '18 at 12:03

what does it mean based on the continuity of the digits?

– Dennis Vash
Nov 24 '18 at 12:12

add a comment |

2 Answers
2

active

oldest

votes

You could sort the array and reduce the array by inserting a new object or updat the end vlaue, depending on the first item or if the last value is not in short range.

In parts:
.sort((a, b) => a - b)
sorts the array by taking the delta of twor elements and returns the needed value for using Array#some. The array fter sorting looks like
[1, 2, 5, 6, 7, 9]
The more complex part is to use Array#reduce which returns the final array of objects.
.reduce((accumulator, value, index) => {

    if (!index || accumulator[accumulator.length - 1].end + 1 < value) {

        accumulator.push({ start: value, end: value });

    } else {

        accumulator[accumulator.length - 1].end = value;

    }

    return accumulator;

}, );
At the beginning, where index is zero, the first object is pushed to the accumulator.

Then it takes the next value form the array and while !1 is not true, the second part of the check
accumulator[accumulator.length - 1].end + 1 < value
is evaluated and returns false, so the else part updates the end property.

Finally the accumulator is returned and contains the wanted result.

var array = [1, 5, 2, 6, 7, 9],

    result = array

        .sort((a, b) => a - b)

        .reduce((accumulator, value, index) => {

            if (!index || accumulator[accumulator.length - 1].end + 1 < value) {

                accumulator.push({ start: value, end: value });

            } else {

                accumulator[accumulator.length - 1].end = value;

            }

            return accumulator;

        }, );

 

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

edited Nov 24 '18 at 12:41

answered Nov 24 '18 at 12:10

Nina Scholz

185k1596168

oh my undertaker it complex answer can you add more info at least i will try to understad

– Hemanth SP
Nov 24 '18 at 12:18

add a comment |

You can use Array.sort, Array.filter and Array.flatMap as well like below

sort will sort the array in increasing order and will result in - [1,2,5,6,7,9]

filter will then filter those results whose value is continuous when checked with left and right values and so it will result in - [1,2,5,7,9] as only 6 is value which has just difference of value 1 when compared with left and right values

flatMap will then loop through the above result and prepare your desired output

let x = [1,5,2,6,7,9]



let res = x.sort()

           .filter((d, i, t) => !(d == t[i-1] + 1 && d == t[i+1] - 1))

           .flatMap((d, i, t) => i%2 == 0 ? [{ st: d, en: t[i+1] || d }] : )



console.log(res)

edited Nov 24 '18 at 13:17

answered Nov 24 '18 at 12:51

Nitish Narang

2,9401815

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

You could sort the array and reduce the array by inserting a new object or updat the end vlaue, depending on the first item or if the last value is not in short range.

In parts:
.sort((a, b) => a - b)
sorts the array by taking the delta of twor elements and returns the needed value for using Array#some. The array fter sorting looks like
[1, 2, 5, 6, 7, 9]
The more complex part is to use Array#reduce which returns the final array of objects.
.reduce((accumulator, value, index) => {

    if (!index || accumulator[accumulator.length - 1].end + 1 < value) {

        accumulator.push({ start: value, end: value });

    } else {

        accumulator[accumulator.length - 1].end = value;

    }

    return accumulator;

}, );
At the beginning, where index is zero, the first object is pushed to the accumulator.

Then it takes the next value form the array and while !1 is not true, the second part of the check
accumulator[accumulator.length - 1].end + 1 < value
is evaluated and returns false, so the else part updates the end property.

Finally the accumulator is returned and contains the wanted result.

var array = [1, 5, 2, 6, 7, 9],

    result = array

        .sort((a, b) => a - b)

        .reduce((accumulator, value, index) => {

            if (!index || accumulator[accumulator.length - 1].end + 1 < value) {

                accumulator.push({ start: value, end: value });

            } else {

                accumulator[accumulator.length - 1].end = value;

            }

            return accumulator;

        }, );

 

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

edited Nov 24 '18 at 12:41

answered Nov 24 '18 at 12:10

Nina Scholz

185k1596168

oh my undertaker it complex answer can you add more info at least i will try to understad

– Hemanth SP
Nov 24 '18 at 12:18

add a comment |

You could sort the array and reduce the array by inserting a new object or updat the end vlaue, depending on the first item or if the last value is not in short range.

In parts:
.sort((a, b) => a - b)
sorts the array by taking the delta of twor elements and returns the needed value for using Array#some. The array fter sorting looks like
[1, 2, 5, 6, 7, 9]
The more complex part is to use Array#reduce which returns the final array of objects.
.reduce((accumulator, value, index) => {

    if (!index || accumulator[accumulator.length - 1].end + 1 < value) {

        accumulator.push({ start: value, end: value });

    } else {

        accumulator[accumulator.length - 1].end = value;

    }

    return accumulator;

}, );
At the beginning, where index is zero, the first object is pushed to the accumulator.

Then it takes the next value form the array and while !1 is not true, the second part of the check
accumulator[accumulator.length - 1].end + 1 < value
is evaluated and returns false, so the else part updates the end property.

Finally the accumulator is returned and contains the wanted result.

var array = [1, 5, 2, 6, 7, 9],

    result = array

        .sort((a, b) => a - b)

        .reduce((accumulator, value, index) => {

            if (!index || accumulator[accumulator.length - 1].end + 1 < value) {

                accumulator.push({ start: value, end: value });

            } else {

                accumulator[accumulator.length - 1].end = value;

            }

            return accumulator;

        }, );

 

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

edited Nov 24 '18 at 12:41

answered Nov 24 '18 at 12:10

Nina Scholz

185k1596168

oh my undertaker it complex answer can you add more info at least i will try to understad

– Hemanth SP
Nov 24 '18 at 12:18

add a comment |

You could sort the array and reduce the array by inserting a new object or updat the end vlaue, depending on the first item or if the last value is not in short range.

In parts:
.sort((a, b) => a - b)
sorts the array by taking the delta of twor elements and returns the needed value for using Array#some. The array fter sorting looks like
[1, 2, 5, 6, 7, 9]
The more complex part is to use Array#reduce which returns the final array of objects.
.reduce((accumulator, value, index) => {

    if (!index || accumulator[accumulator.length - 1].end + 1 < value) {

        accumulator.push({ start: value, end: value });

    } else {

        accumulator[accumulator.length - 1].end = value;

    }

    return accumulator;

}, );
At the beginning, where index is zero, the first object is pushed to the accumulator.

Then it takes the next value form the array and while !1 is not true, the second part of the check
accumulator[accumulator.length - 1].end + 1 < value
is evaluated and returns false, so the else part updates the end property.

Finally the accumulator is returned and contains the wanted result.

var array = [1, 5, 2, 6, 7, 9],

    result = array

        .sort((a, b) => a - b)

        .reduce((accumulator, value, index) => {

            if (!index || accumulator[accumulator.length - 1].end + 1 < value) {

                accumulator.push({ start: value, end: value });

            } else {

                accumulator[accumulator.length - 1].end = value;

            }

            return accumulator;

        }, );

 

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

edited Nov 24 '18 at 12:41

answered Nov 24 '18 at 12:10

Nina Scholz

185k1596168

You could sort the array and reduce the array by inserting a new object or updat the end vlaue, depending on the first item or if the last value is not in short range.

In parts:
.sort((a, b) => a - b)
sorts the array by taking the delta of twor elements and returns the needed value for using Array#some. The array fter sorting looks like
[1, 2, 5, 6, 7, 9]
The more complex part is to use Array#reduce which returns the final array of objects.
.reduce((accumulator, value, index) => {

    if (!index || accumulator[accumulator.length - 1].end + 1 < value) {

        accumulator.push({ start: value, end: value });

    } else {

        accumulator[accumulator.length - 1].end = value;

    }

    return accumulator;

}, );
At the beginning, where index is zero, the first object is pushed to the accumulator.

Then it takes the next value form the array and while !1 is not true, the second part of the check
accumulator[accumulator.length - 1].end + 1 < value
is evaluated and returns false, so the else part updates the end property.

Finally the accumulator is returned and contains the wanted result.

var array = [1, 5, 2, 6, 7, 9],

    result = array

        .sort((a, b) => a - b)

        .reduce((accumulator, value, index) => {

            if (!index || accumulator[accumulator.length - 1].end + 1 < value) {

                accumulator.push({ start: value, end: value });

            } else {

                accumulator[accumulator.length - 1].end = value;

            }

            return accumulator;

        }, );

 

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

var array = [1, 5, 2, 6, 7, 9],

    result = array

        .sort((a, b) => a - b)

        .reduce((accumulator, value, index) => {

            if (!index || accumulator[accumulator.length - 1].end + 1 < value) {

                accumulator.push({ start: value, end: value });

            } else {

                accumulator[accumulator.length - 1].end = value;

            }

            return accumulator;

        }, );

 

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

var array = [1, 5, 2, 6, 7, 9],

    result = array

        .sort((a, b) => a - b)

        .reduce((accumulator, value, index) => {

            if (!index || accumulator[accumulator.length - 1].end + 1 < value) {

                accumulator.push({ start: value, end: value });

            } else {

                accumulator[accumulator.length - 1].end = value;

            }

            return accumulator;

        }, );

 

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

edited Nov 24 '18 at 12:41

answered Nov 24 '18 at 12:10

Nina Scholz

185k1596168

edited Nov 24 '18 at 12:41

answered Nov 24 '18 at 12:10

Nina Scholz

185k1596168

answered Nov 24 '18 at 12:10

Nina Scholz

185k1596168

answered Nov 24 '18 at 12:10

Nina Scholz

185k1596168

oh my undertaker it complex answer can you add more info at least i will try to understad

– Hemanth SP
Nov 24 '18 at 12:18

add a comment |

oh my undertaker it complex answer can you add more info at least i will try to understad

– Hemanth SP
Nov 24 '18 at 12:18

oh my undertaker it complex answer can you add more info at least i will try to understad

– Hemanth SP
Nov 24 '18 at 12:18

add a comment |

You can use Array.sort, Array.filter and Array.flatMap as well like below

sort will sort the array in increasing order and will result in - [1,2,5,6,7,9]

filter will then filter those results whose value is continuous when checked with left and right values and so it will result in - [1,2,5,7,9] as only 6 is value which has just difference of value 1 when compared with left and right values

flatMap will then loop through the above result and prepare your desired output

let x = [1,5,2,6,7,9]



let res = x.sort()

           .filter((d, i, t) => !(d == t[i-1] + 1 && d == t[i+1] - 1))

           .flatMap((d, i, t) => i%2 == 0 ? [{ st: d, en: t[i+1] || d }] : )



console.log(res)

edited Nov 24 '18 at 13:17

answered Nov 24 '18 at 12:51

Nitish Narang

2,9401815

add a comment |

You can use Array.sort, Array.filter and Array.flatMap as well like below

sort will sort the array in increasing order and will result in - [1,2,5,6,7,9]

filter will then filter those results whose value is continuous when checked with left and right values and so it will result in - [1,2,5,7,9] as only 6 is value which has just difference of value 1 when compared with left and right values

flatMap will then loop through the above result and prepare your desired output

let x = [1,5,2,6,7,9]



let res = x.sort()

           .filter((d, i, t) => !(d == t[i-1] + 1 && d == t[i+1] - 1))

           .flatMap((d, i, t) => i%2 == 0 ? [{ st: d, en: t[i+1] || d }] : )



console.log(res)

edited Nov 24 '18 at 13:17

answered Nov 24 '18 at 12:51

Nitish Narang

2,9401815

add a comment |

You can use Array.sort, Array.filter and Array.flatMap as well like below

sort will sort the array in increasing order and will result in - [1,2,5,6,7,9]

filter will then filter those results whose value is continuous when checked with left and right values and so it will result in - [1,2,5,7,9] as only 6 is value which has just difference of value 1 when compared with left and right values

flatMap will then loop through the above result and prepare your desired output

let x = [1,5,2,6,7,9]



let res = x.sort()

           .filter((d, i, t) => !(d == t[i-1] + 1 && d == t[i+1] - 1))

           .flatMap((d, i, t) => i%2 == 0 ? [{ st: d, en: t[i+1] || d }] : )



console.log(res)

edited Nov 24 '18 at 13:17

answered Nov 24 '18 at 12:51

Nitish Narang

2,9401815

You can use Array.sort, Array.filter and Array.flatMap as well like below

sort will sort the array in increasing order and will result in - [1,2,5,6,7,9]

filter will then filter those results whose value is continuous when checked with left and right values and so it will result in - [1,2,5,7,9] as only 6 is value which has just difference of value 1 when compared with left and right values

flatMap will then loop through the above result and prepare your desired output

let x = [1,5,2,6,7,9]



let res = x.sort()

           .filter((d, i, t) => !(d == t[i-1] + 1 && d == t[i+1] - 1))

           .flatMap((d, i, t) => i%2 == 0 ? [{ st: d, en: t[i+1] || d }] : )



console.log(res)

let x = [1,5,2,6,7,9]



let res = x.sort()

           .filter((d, i, t) => !(d == t[i-1] + 1 && d == t[i+1] - 1))

           .flatMap((d, i, t) => i%2 == 0 ? [{ st: d, en: t[i+1] || d }] : )



console.log(res)

let x = [1,5,2,6,7,9]



let res = x.sort()

           .filter((d, i, t) => !(d == t[i-1] + 1 && d == t[i+1] - 1))

           .flatMap((d, i, t) => i%2 == 0 ? [{ st: d, en: t[i+1] || d }] : )



console.log(res)

edited Nov 24 '18 at 13:17

answered Nov 24 '18 at 12:51

Nitish Narang

2,9401815

edited Nov 24 '18 at 13:17

answered Nov 24 '18 at 12:51

Nitish Narang

2,9401815

answered Nov 24 '18 at 12:51

Nitish Narang

2,9401815

answered Nov 24 '18 at 12:51

Nitish Narang

2,9401815

add a comment |

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