# Standard Two Pointers Solution with Explanation in Comments using C++

• ``````class Solution {
public:
int minSubArrayLen(int s, vector<int>& nums) {
// initilaize two pointers and current sum as zeros
int ptr1 = 0;
int ptr2 = 0;
int cur_sum = 0;
// initialize the arrayLen to the one greater than the size
// of the input array "nums"
int arrayLen = nums.size()+1;
// as long as the last pointer is smaller the size
// we will keep checking
while(ptr1<nums.size()){
// the cur_sum is not enough
// and the second pointer is not greater than
// the size we will add a new number to the cur_sum
if(s>cur_sum && ptr2<nums.size() ){
cur_sum += nums[ptr2];
ptr2++;
// else if the the cur_sum is greater than/or equal
// to the target sum, we checked the length of current "window"
// by taking the different betwee two pointer and add one
// if the curren "window" size is less than the previous one
// we update this size to the smallest one
}else if(s<=cur_sum){
if(arrayLen >(ptr2-ptr1)){
arrayLen = ptr2-ptr1;
}
cur_sum -=nums[ptr1];
ptr1++;
// if the second pointer already reached to the end
// and cur_sum is still less than the target sum
// then there is no way to increase the cur_sum
// so we break out of the while loop.
}else if(ptr2 == nums.size() && cur_sum<s){
break;
}
}
// return the result,
// the question asks us to return zero
// for any impossible cases.
if(arrayLen == nums.size()+1){
return 0;
}else{
return arrayLen;
}
}
};``````

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