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Divide an array into K subarrays with a minimum difference

RENOUNCEMENT:

The described problem looks like a competition task. I do not participate in any of them, I do not know about any ongoing competitions that may cause a problem. If there are any of them, I will close the question to stay honest!

I have a problem: a given array of A values ​​and an integer K, splitting A into exactly K non-overlapping adjacent subarrays in such a way that the difference between subaras with minimum and minimum sums of a submar is minimal. It is allowed to rotate A by any number in any direction.

Consider an example:

Input: A = [5 1 1 1 3 2], K = 3

Output: [5] [1 1 1] [3 2], maximum amount = 5, minimum amount = 3, result = 2

I have partially working code (terribly ugly, my bad, but that does not mean product quality):

#include <climits>
#include <cstdio>
#include <cstring>

const int max_n = 50;
const int max_k = 20;

int deps[max_n];

int max (int x, int y) {
  return x > y ? x : y;
}

int min (int x, int y) {
  return x < y ? x : y;
}

int sum (int a[], int start, int end) {
  int res = 0;
  for (int i = start; i <= end; ++i) res += a[i];

  return res;
}

int k_partitioning(int k, int n, int deps[]) {
  int res = INT_MAX;
  // consider all possible rotations/shifts
  for(int offset = 0; offset < n; ++offset) {
    for(int l_min = 0; l_min < n; ++l_min) {
      for(int r_min = l_min; r_min < n; ++r_min) {
        // check minimal sum subarray
        int min_sum = sum (deps, l_min, r_min);

        int dp[k][n];
        for (int s = 0; s < k; ++s) {
          for (int q = 0; q < n; ++q) {
            dp[s][q] = 0;
          }
        }
        // assuming that current sum is a target sum
        dp[0][r_min-l_min] = min_sum;

        for(int p = 1; p < k; ++p) {
          for(int l_max = 0; l_max < n; ++l_max) {
            for(int r_max = 0; r_max < n; ++r_max) {
              int max_sum = sum(deps, l_max, r_max);

              if (max_sum >= min_sum) dp[p][r_max] = max(dp[p-1][l_max], max_sum);
            } // l_maxs
          } // r_maxs
        } // partitions
        // printing dp

        // skip incorrect partitioning, when not all K partitions were used
        if (dp[k-1][n-1] == 0) continue;

        // update difference
        res = min (res, dp[k-1][n-1] - min_sum);
      } // end min sum seg
    } // start min sum seg
    //break;
  } // cuts
  return res;
}

int main(int argc, char* argv[]) {
  int k = 0;
  scanf("%d", &k);

  int n = 0;
  scanf("%d", &n);

  for (int i = 0; i < n; ++i) {
    scanf("%d", &deps[i]);
  }

  printf ("%d\n", k_partitioning(k, n, deps));

  return 0;
}

: , , , , . : O (K * N ^ 4).

, , . - ?

, :

N = 4, K = 2, A = [6 13 10 2]

. -, "" l_min. -, , dp 0 - , ( , max_value ). , - . .

#include <climits>
#include <cstdio>
#include <cstring>

const int max_value = 200000;
const int max_n = 50;
const int max_k = 20;

int deps[max_n];

int max (int x, int y) {
  return x > y ? x : y;
}

int min (int x, int y) {
  return x < y ? x : y;
}

int sum (int a[], int start, int end) {
  int res = 0;
  for (int i = start; i <= end; ++i) res += a[i];

  return res;
}

int k_partitioning(int k, int n, int deps[]) {
  int res = max_value;

  for(int l_min = 0; l_min < n; ++l_min) {
    for(int r_min = l_min; r_min < n; ++r_min) {
      int min_sum = sum (deps, l_min+1, r_min);

      int dp[k][n];
      for (int s = 0; s < k; ++s) {
        for (int q = 0; q < n; ++q) {
          dp[s][q] = max_value;
        }
      }
      // assuming that current sum is a target sum
      dp[0][r_min-l_min] = min_sum;

      for(int p = 1; p < k; ++p) {
        for(int l_max = 0; l_max < n; ++l_max) {
          for(int r_max = l_max; r_max < n; ++r_max) {
            int max_sum = sum(deps, l_max+1, r_max);

            if (max_sum >= min_sum) dp[p][r_max] = max(dp[p-1][l_max], max_sum);
          } // l_maxs
        } // r_maxs
      } // partitions

      // skip incorrect partitioning, when not all K partitions were used
      if (dp[k-1][n-1] == max_value) continue;

      // update difference
      res = min (res, dp[k-1][n-1] - min_sum);
    } // end min sum seg

    // rotate an array to consider different starting points
    int tmp[n];
    for (int i = 0; i < n; ++i) {
      int new_idx = i + n + 1;

      tmp[new_idx % n] = deps[i];
    }

    for(int i = 0; i < n; ++i) deps[i] = tmp[i];
  } // start min sum seg

  return res;
}

int main(int argc, char* argv[]) {
  int k = 0;
  scanf("%d", &k);

  int n = 0;
  scanf("%d", &n);

  for (int i = 0; i < n; ++i) {
    scanf("%d", &deps[i]);
  }

  printf ("%d\n", k_partitioning(k, n, deps));

  return 0;
}
+4
2

, , !

: , 0. , . DP . .

, . .

, , .

#include <climits>
#include <cstdio>
#include <cstring>

const int max_value = 200000;
const int max_n = 50;
const int max_k = 20;

int deps[max_n];

int max (int x, int y) {
  return x > y ? x : y;
}

int min (int x, int y) {
  return x < y ? x : y;
}

int sum (int a[], int start, int end) {
  int res = 0;

  for (int i = start; i <= end; ++i) res += a[i];

  return res;
}

int k_partitioning(int k, int n, int deps[]) {
  int res = max_value;
  for(int offset = 0; offset < n; ++offset) {
    int l_min = 0;
    for(int r_min = l_min; r_min < n; ++r_min) {
      int min_sum = sum (deps, l_min, r_min);

      int dp[k][n];
      for (int s = 0; s < k; ++s) {
        for (int q = 0; q < n; ++q) {
          dp[s][q] = max_value;
        }
      }
      // assuming that current sum is a target sum
      dp[0][r_min-l_min] = min_sum;

      for(int p = 1; p < k; ++p) {
        for(int l_max = r_min; l_max < n; ++l_max) {
          for(int r_max = l_max; r_max < n; ++r_max) {
            int max_sum = sum(deps, l_max+1, r_max);

            if (max_sum >= min_sum) {
              dp[p][r_max] = min(dp[p][r_max], max(dp[p-1][l_max], max_sum));
            }

          } // l_maxs
        } // r_maxs
      } // partitions

      // skip incorrect partitioning, when not all K partitions were used
      if (dp[k-1][n-1] == max_value) continue;

      // update difference
      res = min (res, dp[k-1][n-1] - min_sum);
    } // end min sum seg
    int tmp[n];
    for (int i = 0; i < n; ++i) {
      int new_idx = i + n - 1;

      tmp[new_idx % n] = deps[i];
    }

    for(int i = 0; i < n; ++i) deps[i] = tmp[i];

  } // start min sum seg
  return res;
}

int main(int argc, char* argv[]) {
  int k = 0;
  scanf("%d", &k);

  int n = 0;
  scanf("%d", &n);

  for (int i = 0; i < n; ++i) {
    scanf("%d", &deps[i]);
  }

  printf ("%d\n", k_partitioning(k, n, deps));

  return 0;
}
+3

, , :)

, k , A[i] (sum A[i-j..i]) , f(k-1, i-j-1), - (low, high), , high, new_interval = (low, sum) , low, new_interval = (sum, high); . ,

i:  0 1 2 3 4 5
A: [5 1 1 1 3 2]

k = 3
i = 3, j = 0
The ordered intervals available for f(3-1, 3-0-1) = f(2,2) are:
  (2,5), (1,6) // These were the sums, (A[1..2], A[0]) and (A[2], A[0..1])
Sum = A[3..3-0] = 1
Update intervals: (2,5) -> (1,5)
                  (1,6) -> (1,6) no change

, k.

:

A: [5 1 1 1 3 2]

K = 1:

  N = 0..5; Intervals: (5,5), (6,6), (7,7), (8,8), (11,11), (13,13)

K = 2:

  N = 0: Intervals: N/A

  N = 1: Intervals: (1,5)

  N = 2: (1,6), (2,5)

    Prune: remove (1,6) since any sum <= 1 would be better paired with (2,5)
           and any sum >= 6 would be better paired with (2,5)

  N = 3: (1,7), (2,6), (3,5)

    Prune: remove (2,6) and (1,7)

  N = 4: (3,8), (4,7), (5,6), (5,6)

    Prune: remove (3,8) and (4,7)

  N = 5: (2,11), (5,8), (6,7)

    Prune: remove (2,11) and (5,8)

k = 2 :

{
  k: 2,
  n: {
    1: (1,5),
    2: (2,5),
    3: (3,5),
    4: (5,6),
    5: (6,7)
  }
}

k = 3 n choose 2 n !

, k = 3:

for k' = 1 to k
  for sum A[i-j..i], for i <- [k'-1..n], j <- [0..i-k'+1]:
    for interval in record[k'-1][i-j-1]: // records are for [k'][n']
      update interval
  prune intervals in k'

k' = 3
  i = 2
    sum = 1, record[2][1] = (1,5) -> no change

  i = 3
    // sums are accumulating right to left starting from A[i]
    sum = 1, record[2][2] = (2,5) -> (1,5)
    sum = 2, record[2][1] = (1,5) -> no change

  i = 4
    sum = 3, record[2][3] = (3,5) -> no change
    sum = 4, record[2][2] = (2,5) -> no change
    sum = 5, record[2][1] = (1,5) -> no change

  i = 5
    sum = 2, record[2][4] = (5,6) -> (2,6)
    sum = 5, record[2][3] = (3,5) -> no change
    sum = 6, record[2][2] = (2,5) -> (2,6)
    sum = 7, record[2][1] = (1,5) -> (1,7)

5 record[2][3] = (3,5), (3,5). . , k = 3

{
  k: 3
  n: {
    2: (1,5), 
    3: (1,5),
    4: (3,5),
    5: (3,5)
  }
}
0

Source: https://habr.com/ru/post/1693772/

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