Integer Partition

Definition

Definition

The Integer Partition Algorithm aims to find all possible ways to partition a given integer into a sum of positive integers. It's a fundamental problem in combinatorics and dynamic programming, often used in various fields like number theory and computer science

Explanation

Select the target integer that you wish to partition. Next, set up a two-dimensional array designated for storing partition counts. Iterate through each integer starting from 1 up to the chosen target integer. For each integer i, systematically go through all potential partition sizes j. Employ dynamic programming principles to continually update the counts of partitions. Upon completion, return the count of partitions available for the target integer, thus completing the integer partitioning algorithm

Guidance

• Start with the target integer, let's call it n
• Create a grid called "partitions" with (n+1) rows and (n+1) columns, and fill it with zeros
• For each row in the grid (from 0 to n)
  • set the value at column 1 to 1
• Now, for each row (from 1 to n) and each column (from 1 to n)
  • If the column index is greater than the row index, copy the value from the diagonal (same row, same column)
  • Otherwise, add the value from the previous column to the value from the row-index difference column
• The result is the intersection of the nth row and nth column in the grid

Tips

• utilize dynamic programming to avoid redundant calculations
• take advantage of memoization to store intermediate results
• ensure proper bounds checking to avoid index out of range errors

Practice

Practice

integerPartition(n):
  partitions = 2D array of size (n+1) x (n+1) filled with 0s
  for i from 0 to n:
    partitions[i][1] = 1
  for i from 1 to n:
    for j from 1 to n:
      if j > i:
        partitions[i][j] = partitions[i][i]
      else:
        partitions[i][j] = partitions[i][j-1] + partitions[i-j][j]
  return partitions[n][n]

Solution

package main

func countPartitions(n int) int {
    dp := make([]int, n+1)
    dp[0] = 1
    for i := 1; i <= n; i++ {
        for j := i; j <= n; j++ {
            dp[j] += dp[j-i]
        }
    }
    return dp[n]
}

public class IntegerPartition {

  public static int countPartitions(int n) {
    int[] dp = new int[n + 1];
    dp[0] = 1;
    for (int i = 1; i <= n; i++) {
      for (int j = i; j <= n; j++) {
        dp[j] += dp[j - i];
      }
    }
    return dp[n];
  }
}

function countPartitions(n) {
  let dp = new Array(n + 1).fill(0);
  dp[0] = 1;
  for (let i = 1; i <= n; i++) {
    for (let j = i; j <= n; j++) {
      dp[j] += dp[j - i];
    }
  }
  return dp[n];
}

fun countPartitions(n: Int): Int {
    val dp = IntArray(n + 1)
    dp[0] = 1
    for (i in 1..n) {
        for (j in i..n) {
            dp[j] += dp[j - i]
        }
    }
    return dp[n]
}

def count_partitions(n):
    dp = [0] * (n + 1)
    dp[0] = 1
    for i in range(1, n + 1):
        for j in range(i, n + 1):
            dp[j] += dp[j - i]
    return dp[n]

fn count_partitions(n: usize) -> usize {
    let mut dp = vec![0; n + 1];
    dp[0] = 1;
    for i in 1..=n {
        for j in i..=n {
            dp[j] += dp[j - i];
        }
    }
    dp[n]
}

function countPartitions(n: number): number {
  const dp: number[] = new Array(n + 1).fill(0);
  dp[0] = 1;
  for (let i = 1; i <= n; i++) {
    for (let j = i; j <= n; j++) {
      dp[j] += dp[j - i];
    }
  }
  return dp[n];
}