// 1282. Group the People Given the Group Size They Belong To #include #include #include /* * there are 'n' people that are split into some unknown number of groups. each * person is labeled with a unique id from 0 to 'n - 1'. you are given an * integer array 'group_sizes' where 'group_sizes[i]' is the size of the group * that person 'i' is in. for example if 'group_sizes[i] = 3', then person 1 * must be in a groiup size of 3. return a list of groups such that each person * 'i' is in a group of size 'group_sizes[i]'. each person should appear in * exactly one group, and every person must be in a group. if there are multiple * answers, return any of them. it is guaranteed that there will be at least one * valid solution for the given input. */ int **groupThePeople(int *group_sizes, int group_sizes_size, int *return_size, int **return_col_size) { int table[group_sizes_size + 1][group_sizes_size + 1], col[group_sizes_size + 1]; memset(table, 0, sizeof(table)); memset(col, 0, sizeof(col)); int **ans = (int **)malloc(sizeof(int *) * group_sizes_size); *return_size = 0; *return_col_size = (int *)malloc(sizeof(int) * group_sizes_size); for (int i = 0; i < group_sizes_size; i++) { table[group_sizes[i]][col[group_sizes[i]]] = i; col[group_sizes[i]]++; } for (int i = 0; i <= group_sizes_size; i++) { int k = 0; while (col[i]) { ans[*return_size] = (int *)malloc(sizeof(int) * i); (*return_col_size)[*return_size] = i; for (int j = 0; j < i; j++) { ans[*return_size][j] = table[i][k++]; col[i]--; } (*return_size)++; } } return ans; } int main() { int gs1[] = {3, 3, 3, 3, 3, 1, 3}, gs2[] = {2, 1, 3, 3, 3, 2}; int **gtp1 = groupThePeople(gs1, 7, NULL, NULL); int **gtp2 = groupThePeople(gs2, 6, NULL, NULL); for (int i = 0; i < 3; i++) for (int j = 0; j < 3; j++) printf("%d ", gtp1[i][j]); printf("\n"); for (int i = 0; i < 3; i++) for (int j = 0; j < 3; j++) printf("%d ", gtp2[i][j]); printf("\n"); }