Add integer partition.

README.md

@@ -31,6 +31,7 @@
   * [Primality Test](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/math/primality-test) (trial division method)
   * [Euclidean Algorithm](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/math/euclidean-algorithm) - calculate the Greatest Common Divisor (GCD)
   * [Least Common Multiple](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/math/least-common-multiple) (LCM)
+  * [Integer Partition](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/math/integer-partition)
 * **Sets**
   * [Cartesian Product](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/cartesian-product) - product of multiple sets
   * [Power Set](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/power-set) - all subsets of the set 
@@ -98,9 +99,9 @@
   * [Longest Increasing subsequence](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/longest-increasing-subsequence)
   * [Shortest Common Supersequence](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/shortest-common-supersequence)
   * [0/1 Knapsack Problem](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/knapsack-problem)
+  * [Integer Partition](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/math/integer-partition)
   * Maximum subarray
   * Maximum sum path
-  * Integer Partition
 * **Backtracking**
 * **Branch & Bound**
 

src/algorithms/math/integer-partition/README.md

@@ -0,0 +1,33 @@
+# Integer Partition
+
+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 + 1
+```
+
+The 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.
+
+![Integer Partition](https://upload.wikimedia.org/wikipedia/commons/d/d8/Ferrer_partitioning_diagrams.svg)
+
+## References
+
+- [Wikipedia](https://en.wikipedia.org/wiki/Partition_(number_theory))

src/algorithms/math/integer-partition/__test__/integerPartition.test.js

@@ -0,0 +1,11 @@
+import integerPartition from '../integerPartition';
+
+describe('integerPartition', () => {
+  it('should partition the number', () => {
+    expect(integerPartition(1)).toBe(1);
+    expect(integerPartition(2)).toBe(2);
+    expect(integerPartition(3)).toBe(3);
+    expect(integerPartition(4)).toBe(5);
+    expect(integerPartition(8)).toBe(22);
+  });
+});

src/algorithms/math/integer-partition/integerPartition.js

@@ -0,0 +1,47 @@
+/**
+ * @param {Number} 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 row. It represents the number of way of how we can form
+  // number zero out of numbers 0, 1, 2, ... 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
+        // previous number but WITH current summand.
+        const combosWithoutSummand = partitionMatrix[summandIndex - 1][numberIndex];
+        const combosWithSummand = partitionMatrix[summandIndex][numberIndex - summandIndex];
+
+        partitionMatrix[summandIndex][numberIndex] = combosWithoutSummand + combosWithSummand;
+      }
+    }
+  }
+
+  return partitionMatrix[number][number];
+}