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+// 2028. Find Missing Observations
+#include "leetcode.h"
+
+/*
+ * you have observations of n + m 6 sided dice rolls withh each face numbered
+ * from 1 to 6. n of the observations went missing and you only have the
+ * observations of m rolls. fortunately, you have also calculated thhe average
+ * value of the n + m rolls. you are given an integer array rolls of length m
+ * where rolls[i] is the value of the i'th observation. you are also given the
+ * two integers mean and n. return an array of length n containing thhe missing
+ * observations such that the average value of the n + m rolls is exactly mean.
+ * if there are multiple valid answers, return any of them. if no such array
+ * exists, return an empty array. thhe average value of a set of k numbers is
+ * the sum of the numbers divided by k. note that mean is an integer so the sum
+ * of the n + m rolls should be divisble by n + m.
+ */
+
+int *missingRolls(int *rolls, int rolls_size, int mean, int n,
+ int *return_size) {
+ int sum = 0;
+ for (int i = 0; i < rolls_size; i++)
+ sum += rolls[i];
+ int target = mean * (rolls_size + n);
+ int diff = target - sum;
+ if (diff < n || diff > 6 * n) {
+ *return_size = 0;
+ return NULL;
+ }
+ int *ans = malloc(n * sizeof(int));
+ *return_size = n;
+ for (int i = 0; i < n; i++)
+ ans[i] = diff / n;
+ diff -= n * (diff / n);
+ int pos = 0;
+ while (diff) {
+ if (diff > (6 - ans[pos])) {
+ diff -= (6 - ans[pos]);
+ ans[pos] = 6;
+ } else {
+ ans[pos] += diff;
+ diff = 0;
+ }
+ pos++;
+ }
+ return ans;
+}
+
+int main() {
+ int r1[] = {3, 2, 4, 3}, r2[] = {1, 5, 6}, r3[] = {1, 2, 3, 4};
+ int rs1[] = {}, rs2[] = {}, rs3[] = {};
+ int *mr1 = missingRolls(r1, 4, 4, 2, rs1),
+ *mr2 = missingRolls(r2, 3, 3, 4, rs2),
+ *mr3 = missingRolls(r3, 4, 6, 4, rs3);
+ for (int i = 0; i < 2; i++)
+ printf("%d ", mr1[i]); // expect: 6 6
+ printf("\n");
+ for (int i = 0; i < 4; i++)
+ printf("%d ", mr2[i]); // expect: 2 3 2 2
+ printf("\n");
+ for (int i = 0; i < 1; i++)
+ printf("%d ", mr3[i]); // expect: null
+ printf("\n");
+}