JavaScript/Dynamic-Programming/RodCutting.js at master · Guduly/JavaScript · GitHub

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/*
 * You are given a rod of 'n' length and an array of prices associated with all the lengths less than 'n'.
 * Find the maximum profit possible by cutting the rod and selling the pieces.
*/

exportfunctionrodCut(prices,n){
constmemo=newArray(n+1)
memo[0]=0

for(leti=1;i<=n;i++){
letmaxVal=Number.MIN_VALUE
for(letj=0;j<i;j++){maxVal=Math.max(maxVal,prices[j]+memo[i-j-1])}
memo[i]=maxVal
}

returnmemo[n]
}
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/*
 * You are given a rod of 'n' length and an array of prices associated with all the lengths less than 'n'.
 * Find the maximum profit possible by cutting the rod and selling the pieces.
*/

exportfunctionrodCut(prices,n){
constmemo=newArray(n+1)
memo[0]=0

for(leti=1;i<=n;i++){
letmaxVal=Number.MIN_VALUE
for(letj=0;j<i;j++){maxVal=Math.max(maxVal,prices[j]+memo[i-j-1])}
memo[i]=maxVal
}

returnmemo[n]
}
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/*
 * You are given a rod of 'n' length and an array of prices associated with all the lengths less than 'n'.
 * Find the maximum profit possible by cutting the rod and selling the pieces.
*/

exportfunctionrodCut(prices,n){
constmemo=newArray(n+1)
memo[0]=0

for(leti=1;i<=n;i++){
letmaxVal=Number.MIN_VALUE
for(letj=0;j<i;j++){maxVal=Math.max(maxVal,prices[j]+memo[i-j-1])}
memo[i]=maxVal
}

returnmemo[n]
}
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/*
 * You are given a rod of 'n' length and an array of prices associated with all the lengths less than 'n'.
 * Find the maximum profit possible by cutting the rod and selling the pieces.
*/

exportfunctionrodCut(prices,n){
constmemo=newArray(n+1)
memo[0]=0

for(leti=1;i<=n;i++){
letmaxVal=Number.MIN_VALUE
for(letj=0;j<i;j++){maxVal=Math.max(maxVal,prices[j]+memo[i-j-1])}
memo[i]=maxVal
}

returnmemo[n]
}
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/*
 * You are given a rod of 'n' length and an array of prices associated with all the lengths less than 'n'.
 * Find the maximum profit possible by cutting the rod and selling the pieces.
*/

exportfunctionrodCut(prices,n){
constmemo=newArray(n+1)
memo[0]=0

for(leti=1;i<=n;i++){
letmaxVal=Number.MIN_VALUE
for(letj=0;j<i;j++){maxVal=Math.max(maxVal,prices[j]+memo[i-j-1])}
memo[i]=maxVal
}

returnmemo[n]
}
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/*
 * You are given a rod of 'n' length and an array of prices associated with all the lengths less than 'n'.
 * Find the maximum profit possible by cutting the rod and selling the pieces.
*/

exportfunctionrodCut(prices,n){
constmemo=newArray(n+1)
memo[0]=0

for(leti=1;i<=n;i++){
letmaxVal=Number.MIN_VALUE
for(letj=0;j<i;j++){maxVal=Math.max(maxVal,prices[j]+memo[i-j-1])}
memo[i]=maxVal
}

returnmemo[n]
}
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/*
 * You are given a rod of 'n' length and an array of prices associated with all the lengths less than 'n'.
 * Find the maximum profit possible by cutting the rod and selling the pieces.
*/

exportfunctionrodCut(prices,n){
constmemo=newArray(n+1)
memo[0]=0

for(leti=1;i<=n;i++){
letmaxVal=Number.MIN_VALUE
for(letj=0;j<i;j++){maxVal=Math.max(maxVal,prices[j]+memo[i-j-1])}
memo[i]=maxVal
}

returnmemo[n]
}
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/*
 * You are given a rod of 'n' length and an array of prices associated with all the lengths less than 'n'.
 * Find the maximum profit possible by cutting the rod and selling the pieces.
*/

exportfunctionrodCut(prices,n){
constmemo=newArray(n+1)
memo[0]=0

for(leti=1;i<=n;i++){
letmaxVal=Number.MIN_VALUE
for(letj=0;j<i;j++){maxVal=Math.max(maxVal,prices[j]+memo[i-j-1])}
memo[i]=maxVal
}

returnmemo[n]
}