1764. Form Array by Concatenating Subarrays of Another Array
Explanation:
To solve this problem, we need to check if it is possible to choose n disjoint subarrays from the nums array such that they match the groups array and are in the same order as specified by groups.
We can iterate through the groups array and for each group, try to match it with a subarray in nums. We need to ensure that the subarrays are disjoint and appear in the same order as in groups.
Algorithmic Idea:
- Initialize a pointer
jto keep track of the current position in thenumsarray. - Iterate over each group in
groups: a. Find the subarray innumsthat matches the current group. b. If found, update the pointerjto the next position after the matched subarray. c. If not found, return false. - Return true if all groups are successfully matched.
Time Complexity:
The time complexity of this algorithm is O(n * m), where n is the number of groups and m is the maximum length of any group.
Space Complexity:
The space complexity of this algorithm is O(1) as we are using constant extra space.
:
class Solution {
public boolean canChoose(int[][] groups, int[] nums) {
int j = 0;
for (int[] group : groups) {
int i = j;
while (i < nums.length && !matchGroup(group, nums, i)) {
i++;
}
if (i >= nums.length) {
return false;
}
j = i + group.length;
}
return true;
}
private boolean matchGroup(int[] group, int[] nums, int start) {
for (int i = 0; i < group.length; i++) {
if (start + i >= nums.length || group[i] != nums[start + i]) {
return false;
}
}
return true;
}
}Code Editor (Testing phase)
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