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Missing Ranges in C++
Suppose we have a sorted integer array nums, the range of elements are in the inclusive range [lower, upper], we have to find the missing ranges.
So, if the input is like nums = [0, 1, 3, 50, 75], and lower value is 0 and upper value is 99, then the output will be ["2", "4->49", "51->74", "76->99"]
To solve this, we will follow these steps −
Define an array nums
Define one set v
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for initialize i := 0, when i < size of t, update (increase i by 1), do −
-
if t[i] is not in v, then −
insert t[i] into v
insert t[i] at the end of nums
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define one array called ret
curr := lower
i := 0, n := size of nums
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while curr <= upper, do −
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if i < n and nums[i] is same as curr, then −
(increase i by 1)
(increase curr by 1)
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Otherwise
temp := convert curr to string
(increase curr by 1)
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if i < n and nums[i] is same as curr, then −
insert temp at the end of ret
Ignore following part, skip to the next iteration
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Otherwise
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if i is same as n, then −
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if curr <= upper, then −
temp := "->"
temp := temp concatenate upper as string
curr := upper + 1
insert temp at the end of ret
-
-
Otherwise
temp := "->"
curr := nums[i]
temp := temp concatenate (curr - 1) as string
curr := nums[i]
insert temp at the end of ret
-
-
return ret
Example
Let us see the following implementation to get a better understanding −
#include <bits/stdc++.h> using namespace std; void print_vector(vector<auto< v){ cout << "["; for(int i = 0; i<v.size(); i++){ cout << v[i] << ", "; } cout << "]"<<endl; } class Solution { public: vector<string< findMissingRanges(vector<int<& t, int lower, int upper) { vector <int< nums; set <long long int> v; for(int i = 0; i < t.size(); i++){ if(!v.count(t[i])){ v.insert(t[i]); nums.push_back(t[i]); } } vector < string > ret; long long int curr = lower; int i = 0; int n = nums.size(); while(curr <= upper){ if(i < n && nums[i] == curr){ i++; curr++; } else{ string temp = to_string(curr); curr++; if(i < n && nums[i] == curr){ ret.push_back(temp); continue; } else{ if(i == n){ if(curr <= upper){ temp += "->"; temp += to_string(upper); curr = (long long int )upper + 1; } ret.push_back(temp); } else{ temp += "->"; curr = nums[i]; temp += to_string(curr - 1); curr = nums[i]; ret.push_back(temp); } } } } return ret; } }; main(){ Solution ob; vector<int< v = {0,1,3,50,75}; print_vector(ob.findMissingRanges(v, 0, 99)); }
Input
{0,1,3,50,75}, 0, 99
Output
[2, 4->49, 51->74, 76->99, ]