Find the minimum and maximum values in the nested object
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0
down vote
favorite
I have a deeply nested javascript object with an unlimited amout of children. Every child has a value.
var object = {
value: 1,
children: {
value: 10,
children:{
value: 2,
children: {...}
}
}
}
All attempts to make a recursive function did not succeed, it turned out to go down only to a lower level.
javascript reduce
asked Nov 19 at 16:38
LuisDemn
42
add a comment |
up vote
0
down vote
favorite
I have a deeply nested javascript object with an unlimited amout of children. Every child has a value.
var object = {
value: 1,
children: {
value: 10,
children:{
value: 2,
children: {...}
}
}
}
All attempts to make a recursive function did not succeed, it turned out to go down only to a lower level.
javascript reduce
asked Nov 19 at 16:38
LuisDemn
42
3
What was your previous unsuccessful attempt?
– Robin Zigmond
Nov 19 at 16:39
Reopening. The linked question is quite different: no recursion is needed in that one..
– Scott Sauyet
Nov 19 at 17:00
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a deeply nested javascript object with an unlimited amout of children. Every child has a value.
var object = {
value: 1,
children: {
value: 10,
children:{
value: 2,
children: {...}
}
}
}
All attempts to make a recursive function did not succeed, it turned out to go down only to a lower level.
javascript reduce
asked Nov 19 at 16:38
LuisDemn
42
I have a deeply nested javascript object with an unlimited amout of children. Every child has a value.
var object = {
value: 1,
children: {
value: 10,
children:{
value: 2,
children: {...}
}
}
}
All attempts to make a recursive function did not succeed, it turned out to go down only to a lower level.
javascript reduce
javascript reduce
asked Nov 19 at 16:38
LuisDemn
42
asked Nov 19 at 16:38
LuisDemn
42
asked Nov 19 at 16:38
LuisDemn
42
asked Nov 19 at 16:38
LuisDemn
42
asked Nov 19 at 16:38
LuisDemn
42
42
3
What was your previous unsuccessful attempt?
– Robin Zigmond
Nov 19 at 16:39
Reopening. The linked question is quite different: no recursion is needed in that one..
– Scott Sauyet
Nov 19 at 17:00
add a comment |
3
What was your previous unsuccessful attempt?
– Robin Zigmond
Nov 19 at 16:39
Reopening. The linked question is quite different: no recursion is needed in that one..
– Scott Sauyet
Nov 19 at 17:00
3
3
What was your previous unsuccessful attempt?
– Robin Zigmond
Nov 19 at 16:39
What was your previous unsuccessful attempt?
– Robin Zigmond
Nov 19 at 16:39
Reopening. The linked question is quite different: no recursion is needed in that one..
– Scott Sauyet
Nov 19 at 17:00
Reopening. The linked question is quite different: no recursion is needed in that one..
– Scott Sauyet
Nov 19 at 17:00
add a comment |
3 Answers
3
active
oldest
votes
up vote
3
down vote
After flattening your linked list into an array, you can use Array.prototype.reduce() with an accumulator that is a tuple of min and max, starting with initial values of Infinity and -Infinity respectively to match the implementations of Math.min() and Math.max():
const object = {
value: 1,
children: {
value: 10,
children: {
value: 2,
children: {
value: 5,
children: null
}
}
}
}
const flat = o => o == null || o.value == null ? : [o.value, ...flat(o.children)]
const [min, max] = flat(object).reduce(
([min, max], value) => [Math.min(min, value), Math.max(max, value)],
[Infinity, -Infinity]
)
console.log(min, max)answered Nov 19 at 16:50
Patrick Roberts
18.3k23372
add a comment |
up vote
2
down vote
Since children is an object with only one value (vs an array with potentially many), this is a pretty simple recursive function. The base case is when there are no children in which case both min and max are just the value. Otherwise recurse on the children to find the min and max:
var object = {
value: -10,
children: {
value: 4,
children:{
value: 200,
children: {
value: -100,
children: null
}
}
}
}
function getMinMax(obj) {
if (!obj.children || obj.children.value == undefined)
return {min: obj.value, max: obj.value}
else {
let m = getMinMax(obj.children)
return {min: Math.min(obj.value, m.min), max: Math.max(obj.value, m.max)}
}
}
console.log(getMinMax(object))answered Nov 19 at 16:51
Mark Meyer
30.6k32550
add a comment |
up vote
1
down vote
Short and simple, for min change Math.max to Math.min
var test = {
value: 1,
children: {
value: 10,
children:{
value: 2,
children: {}
}
}
}
function findMaxValue(obj) {
if (Object.keys(obj.children).length === 0) {
return obj.value;
}
return Math.max(obj.value, findMaxValue(obj.children))
}
console.log(findMaxValue(test))answered Nov 19 at 16:51
dotconnor
87416
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
After flattening your linked list into an array, you can use Array.prototype.reduce() with an accumulator that is a tuple of min and max, starting with initial values of Infinity and -Infinity respectively to match the implementations of Math.min() and Math.max():
const object = {
value: 1,
children: {
value: 10,
children: {
value: 2,
children: {
value: 5,
children: null
}
}
}
}
const flat = o => o == null || o.value == null ? : [o.value, ...flat(o.children)]
const [min, max] = flat(object).reduce(
([min, max], value) => [Math.min(min, value), Math.max(max, value)],
[Infinity, -Infinity]
)
console.log(min, max)answered Nov 19 at 16:50
Patrick Roberts
18.3k23372
add a comment |
up vote
3
down vote
After flattening your linked list into an array, you can use Array.prototype.reduce() with an accumulator that is a tuple of min and max, starting with initial values of Infinity and -Infinity respectively to match the implementations of Math.min() and Math.max():
const object = {
value: 1,
children: {
value: 10,
children: {
value: 2,
children: {
value: 5,
children: null
}
}
}
}
const flat = o => o == null || o.value == null ? : [o.value, ...flat(o.children)]
const [min, max] = flat(object).reduce(
([min, max], value) => [Math.min(min, value), Math.max(max, value)],
[Infinity, -Infinity]
)
console.log(min, max)answered Nov 19 at 16:50
Patrick Roberts
18.3k23372
add a comment |
up vote
3
down vote
up vote
3
down vote
After flattening your linked list into an array, you can use Array.prototype.reduce() with an accumulator that is a tuple of min and max, starting with initial values of Infinity and -Infinity respectively to match the implementations of Math.min() and Math.max():
const object = {
value: 1,
children: {
value: 10,
children: {
value: 2,
children: {
value: 5,
children: null
}
}
}
}
const flat = o => o == null || o.value == null ? : [o.value, ...flat(o.children)]
const [min, max] = flat(object).reduce(
([min, max], value) => [Math.min(min, value), Math.max(max, value)],
[Infinity, -Infinity]
)
console.log(min, max)answered Nov 19 at 16:50
Patrick Roberts
18.3k23372
After flattening your linked list into an array, you can use Array.prototype.reduce() with an accumulator that is a tuple of min and max, starting with initial values of Infinity and -Infinity respectively to match the implementations of Math.min() and Math.max():
const object = {
value: 1,
children: {
value: 10,
children: {
value: 2,
children: {
value: 5,
children: null
}
}
}
}
const flat = o => o == null || o.value == null ? : [o.value, ...flat(o.children)]
const [min, max] = flat(object).reduce(
([min, max], value) => [Math.min(min, value), Math.max(max, value)],
[Infinity, -Infinity]
)
console.log(min, max)const object = {
value: 1,
children: {
value: 10,
children: {
value: 2,
children: {
value: 5,
children: null
}
}
}
}
const flat = o => o == null || o.value == null ? : [o.value, ...flat(o.children)]
const [min, max] = flat(object).reduce(
([min, max], value) => [Math.min(min, value), Math.max(max, value)],
[Infinity, -Infinity]
)
console.log(min, max)const object = {
value: 1,
children: {
value: 10,
children: {
value: 2,
children: {
value: 5,
children: null
}
}
}
}
const flat = o => o == null || o.value == null ? : [o.value, ...flat(o.children)]
const [min, max] = flat(object).reduce(
([min, max], value) => [Math.min(min, value), Math.max(max, value)],
[Infinity, -Infinity]
)
console.log(min, max)answered Nov 19 at 16:50
Patrick Roberts
18.3k23372
answered Nov 19 at 16:50
Patrick Roberts
18.3k23372
answered Nov 19 at 16:50
Patrick Roberts
18.3k23372
answered Nov 19 at 16:50
Patrick Roberts
18.3k23372
18.3k23372
add a comment |
add a comment |
up vote
2
down vote
Since children is an object with only one value (vs an array with potentially many), this is a pretty simple recursive function. The base case is when there are no children in which case both min and max are just the value. Otherwise recurse on the children to find the min and max:
var object = {
value: -10,
children: {
value: 4,
children:{
value: 200,
children: {
value: -100,
children: null
}
}
}
}
function getMinMax(obj) {
if (!obj.children || obj.children.value == undefined)
return {min: obj.value, max: obj.value}
else {
let m = getMinMax(obj.children)
return {min: Math.min(obj.value, m.min), max: Math.max(obj.value, m.max)}
}
}
console.log(getMinMax(object))answered Nov 19 at 16:51
Mark Meyer
30.6k32550
add a comment |
up vote
2
down vote
Since children is an object with only one value (vs an array with potentially many), this is a pretty simple recursive function. The base case is when there are no children in which case both min and max are just the value. Otherwise recurse on the children to find the min and max:
var object = {
value: -10,
children: {
value: 4,
children:{
value: 200,
children: {
value: -100,
children: null
}
}
}
}
function getMinMax(obj) {
if (!obj.children || obj.children.value == undefined)
return {min: obj.value, max: obj.value}
else {
let m = getMinMax(obj.children)
return {min: Math.min(obj.value, m.min), max: Math.max(obj.value, m.max)}
}
}
console.log(getMinMax(object))answered Nov 19 at 16:51
Mark Meyer
30.6k32550
add a comment |
up vote
2
down vote
up vote
2
down vote
Since children is an object with only one value (vs an array with potentially many), this is a pretty simple recursive function. The base case is when there are no children in which case both min and max are just the value. Otherwise recurse on the children to find the min and max:
var object = {
value: -10,
children: {
value: 4,
children:{
value: 200,
children: {
value: -100,
children: null
}
}
}
}
function getMinMax(obj) {
if (!obj.children || obj.children.value == undefined)
return {min: obj.value, max: obj.value}
else {
let m = getMinMax(obj.children)
return {min: Math.min(obj.value, m.min), max: Math.max(obj.value, m.max)}
}
}
console.log(getMinMax(object))answered Nov 19 at 16:51
Mark Meyer
30.6k32550
Since children is an object with only one value (vs an array with potentially many), this is a pretty simple recursive function. The base case is when there are no children in which case both min and max are just the value. Otherwise recurse on the children to find the min and max:
var object = {
value: -10,
children: {
value: 4,
children:{
value: 200,
children: {
value: -100,
children: null
}
}
}
}
function getMinMax(obj) {
if (!obj.children || obj.children.value == undefined)
return {min: obj.value, max: obj.value}
else {
let m = getMinMax(obj.children)
return {min: Math.min(obj.value, m.min), max: Math.max(obj.value, m.max)}
}
}
console.log(getMinMax(object))var object = {
value: -10,
children: {
value: 4,
children:{
value: 200,
children: {
value: -100,
children: null
}
}
}
}
function getMinMax(obj) {
if (!obj.children || obj.children.value == undefined)
return {min: obj.value, max: obj.value}
else {
let m = getMinMax(obj.children)
return {min: Math.min(obj.value, m.min), max: Math.max(obj.value, m.max)}
}
}
console.log(getMinMax(object))var object = {
value: -10,
children: {
value: 4,
children:{
value: 200,
children: {
value: -100,
children: null
}
}
}
}
function getMinMax(obj) {
if (!obj.children || obj.children.value == undefined)
return {min: obj.value, max: obj.value}
else {
let m = getMinMax(obj.children)
return {min: Math.min(obj.value, m.min), max: Math.max(obj.value, m.max)}
}
}
console.log(getMinMax(object))answered Nov 19 at 16:51
Mark Meyer
30.6k32550
edited Nov 19 at 17:18
answered Nov 19 at 16:51
Mark Meyer
30.6k32550
answered Nov 19 at 16:51
Mark Meyer
30.6k32550
answered Nov 19 at 16:51
Mark Meyer
30.6k32550
30.6k32550
add a comment |
add a comment |
up vote
1
down vote
Short and simple, for min change Math.max to Math.min
var test = {
value: 1,
children: {
value: 10,
children:{
value: 2,
children: {}
}
}
}
function findMaxValue(obj) {
if (Object.keys(obj.children).length === 0) {
return obj.value;
}
return Math.max(obj.value, findMaxValue(obj.children))
}
console.log(findMaxValue(test))answered Nov 19 at 16:51
dotconnor
87416
add a comment |
up vote
1
down vote
Short and simple, for min change Math.max to Math.min
var test = {
value: 1,
children: {
value: 10,
children:{
value: 2,
children: {}
}
}
}
function findMaxValue(obj) {
if (Object.keys(obj.children).length === 0) {
return obj.value;
}
return Math.max(obj.value, findMaxValue(obj.children))
}
console.log(findMaxValue(test))answered Nov 19 at 16:51
dotconnor
87416
add a comment |
up vote
1
down vote
up vote
1
down vote
Short and simple, for min change Math.max to Math.min
var test = {
value: 1,
children: {
value: 10,
children:{
value: 2,
children: {}
}
}
}
function findMaxValue(obj) {
if (Object.keys(obj.children).length === 0) {
return obj.value;
}
return Math.max(obj.value, findMaxValue(obj.children))
}
console.log(findMaxValue(test))answered Nov 19 at 16:51
dotconnor
87416
Short and simple, for min change Math.max to Math.min
var test = {
value: 1,
children: {
value: 10,
children:{
value: 2,
children: {}
}
}
}
function findMaxValue(obj) {
if (Object.keys(obj.children).length === 0) {
return obj.value;
}
return Math.max(obj.value, findMaxValue(obj.children))
}
console.log(findMaxValue(test))var test = {
value: 1,
children: {
value: 10,
children:{
value: 2,
children: {}
}
}
}
function findMaxValue(obj) {
if (Object.keys(obj.children).length === 0) {
return obj.value;
}
return Math.max(obj.value, findMaxValue(obj.children))
}
console.log(findMaxValue(test))var test = {
value: 1,
children: {
value: 10,
children:{
value: 2,
children: {}
}
}
}
function findMaxValue(obj) {
if (Object.keys(obj.children).length === 0) {
return obj.value;
}
return Math.max(obj.value, findMaxValue(obj.children))
}
console.log(findMaxValue(test))answered Nov 19 at 16:51
dotconnor
87416
answered Nov 19 at 16:51
dotconnor
87416
answered Nov 19 at 16:51
dotconnor
87416
answered Nov 19 at 16:51
dotconnor
87416
87416
add a comment |
add a comment |
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3
What was your previous unsuccessful attempt?
– Robin Zigmond
Nov 19 at 16:39
Reopening. The linked question is quite different: no recursion is needed in that one..
– Scott Sauyet
Nov 19 at 17:00