Integer Partition
Algorithm · Stretch
Step through it: Dynamic Programming pattern
Step 1 of 6
Goal: how many ways can you climb 5 stairs taking 1 or 2 steps at a time? You reach stair 5 from stair 4 or stair 3, so ways(5) = ways(4) + ways(3). Working that out naively repeats the same sub-questions many times. Instead write each answer down once.
Step 2 of 6
Base cases: there is 1 way to be on stair 0 (do nothing) and 1 way to reach stair 1.
Step 3 of 6
Stair 2 is reached from stair 1 or stair 0.
Step 4 of 6
Stair 3 comes from stairs 2 and 1.
Step 5 of 6
Stair 4: 3 + 2 = 5. Stair 5: 5 + 3 = 8. Each answer reuses two stored answers, so n stairs take n steps instead of exponential time. Storing sub-answers like this is dynamic programming.
Step 6 of 6
Your turn.
In number theory and combinatorics, a partition of a positive
integer n, also called an integer partition, is a way of
writing n as a sum of positive integers.
Two sums that differ only in the order of their summands are
considered the same partition. For example, 4 can be partitioned
in five distinct ways:
4
3 + 1
2 + 2
2 + 1 + 1
1 + 1 + 1 + 1The order-dependent composition 1 + 3 is the same partition
as 3 + 1, while the two distinct
compositions 1 + 2 + 1 and 1 + 1 + 2 represent the same
partition 2 + 1 + 1.
Young diagrams associated to the partitions of the positive
integers 1 through 8. They are arranged so that images
under the reflection about the main diagonal of the square
are conjugate partitions.
Code
/**
* @param {number} number
* @return {number}
*/
export default function integerPartition(number) {
// Create partition matrix for solving this task using Dynamic Programming.
const partitionMatrix = Array(number + 1).fill(null).map(() => {
return Array(number + 1).fill(null);
});
// Fill partition matrix with initial values.
// Let's fill the first row that represents how many ways we would have
// to combine the numbers 1, 2, 3, ..., n with number 0. We would have zero
// ways obviously since with zero number we may form only zero.
for (let numberIndex = 1; numberIndex <= number; numberIndex += 1) {
partitionMatrix[0][numberIndex] = 0;
}
// Let's fill the first column. It represents the number of ways we can form
// number zero out of numbers 0, 0 and 1, 0 and 1 and 2, 0 and 1 and 2 and 3, ...
// Obviously there is only one way we could form number 0
// and it is with number 0 itself.
for (let summandIndex = 0; summandIndex <= number; summandIndex += 1) {
partitionMatrix[summandIndex][0] = 1;
}
// Now let's go through other possible options of how we could form number m out of
// summands 0, 1, ..., m using Dynamic Programming approach.
for (let summandIndex = 1; summandIndex <= number; summandIndex += 1) {
for (let numberIndex = 1; numberIndex <= number; numberIndex += 1) {
if (summandIndex > numberIndex) {
// If summand number is bigger then current number itself then just it won't add
// any new ways of forming the number. Thus we may just copy the number from row above.
partitionMatrix[summandIndex][numberIndex] = partitionMatrix[summandIndex - 1][numberIndex];
} else {
/*
* The number of combinations would equal to number of combinations of forming the same
* number but WITHOUT current summand number PLUS number of combinations of forming the
* <current number - current summand> number but WITH current summand.
*
* Example:
* Number of ways to form 5 using summands {0, 1, 2} would equal the SUM of:
* - number of ways to form 5 using summands {0, 1} (we've excluded summand 2)
* - number of ways to form 3 (because 5 - 2 = 3) using summands {0, 1, 2}
* (we've included summand 2)
*/
const combosWithoutSummand = partitionMatrix[summandIndex - 1][numberIndex];
const combosWithSummand = partitionMatrix[summandIndex][numberIndex - summandIndex];
partitionMatrix[summandIndex][numberIndex] = combosWithoutSummand + combosWithSummand;
}
}
}
return partitionMatrix[number][number];
}References
Practice
Dynamic Programming
- Climbing StairsEasy
- Fibonacci NumberEasy
- Min Cost Climbing StairsEasy
- Pascal's TriangleEasy
- House RobberMedium
- House Robber IIMedium
- Coin ChangeMedium
- Coin Change IIMedium
- Longest Increasing SubsequenceMedium
- Longest Common SubsequenceMedium
- Word BreakMedium
- Unique PathsMedium
- Unique Paths IIMedium
- Minimum Path SumMedium
- Jump GameMedium
- Partition Equal Subset SumMedium
- Target SumMedium
- Decode WaysMedium
- Maximum Product SubarrayMedium
- Longest Palindromic SubstringMedium
- Palindromic SubstringsMedium
- Best Time to Buy and Sell Stock with CooldownMedium
- TriangleMedium
- Maximal SquareMedium
- Edit DistanceHard
- Regular Expression MatchingHard
- Burst BalloonsHard
- Distinct SubsequencesHard
- Longest Valid ParenthesesHard
- Best Time to Buy and Sell Stock IVHard
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