Corrected explanations and included an example (trekhleb#75)

Author: Nnadozie

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

@@ -17,9 +17,10 @@ export default function integerPartition(number) {
     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.
+  // 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;
   }
@@ -35,7 +36,18 @@ export default function integerPartition(number) {
       } 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.
+        // <current number - current summand> number but WITH current summand.
+        // Example: number of ways to form number 4 using summands 1, 2 and 3 is the sum of
+        // {number of ways to form 4 with sums that begin with 1 +
+        // number of ways to form 4 with sums that begin with 2 and include 1} +
+        // {number of ways to form 4 with sums that begin with 3 and include 2 and 1}
+        // Taking these sums to proceed in descending order of intergers, this gives us:
+        // With 1: 1+1+1+1  -> 1 way
+        // With 2: 2+2, 2+1+1 -> 2 ways
+        // With 3: 3 + (4-3) <= convince yourself that number of ways to form 4 starting
+        // with 3 is == number of ways to form 4-3 where 4-3 == <current number-current summand>
+        // Helper: if there are n ways to get (4-3) then 4 can be represented as 3 + first way,
+        // 3 + second way, and so on until the 3 + nth way. So answer for 4 is: 1 + 2 + 1 = 4 ways
         const combosWithoutSummand = partitionMatrix[summandIndex - 1][numberIndex];
         const combosWithSummand = partitionMatrix[summandIndex][numberIndex - summandIndex];