First missing Positive | Programming Interview Preparation Must Do programming questions
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- čas přidán 18. 11. 2022
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there is one edge case it won't work with duplicate value ex - [1, 1] you will mark 1 by -1 then you will go to second index then you will again do -1 * -1 then it will become 1 but we have to return 2 ans for that we can write one edge case if(arr[index]>0) {arr[index] *= -1 ; }
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It will be nums[index]=-1*abs(nums[index])
this code does not work in gfg !! fail to satisfy all test case .
it won't work for duplicate as well as for negative.
Very helpful videos!
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Amazing concept 🤩
Minor fix required for duplicate items ( [2,3,1,1] ) its returns wrong answer, so fix is that
if(idx < n && nums[idx] > 0) {
nums[idx] *= -1;
}
Air gfg me teat case fail ho raha hai
test case wrong aayenge first sort the array sort(arr,arr+n); then iterate
nums =[1,1]. for this case it will fail.....
haa leetcode pe
int firstMissingPositive(vector& nums) {
for(int i=0;i
bhaiya it will fail for [1,1]
class Solution {
public:
int firstMissingPositive(vector& nums) {
for(int i = 0;i < nums.size(); i++){
if(nums[i] 0){
nums[index]*=-1;
}
}
for(int i = 0;i < nums.size();i++){
if(nums[i] > 0){
return i+1;
}
}
return nums.size()+1;
}
};
bhaiya it will fail on test case 1,1
Why
@@HelloWorldbyprince check on leetcode plz
@@HelloWorldbyprince int firstMissingPositive(vector& nums) {
for(int i=0;i
int firstMissingPositive(vector& nums) {
for(int i=0;i
It will fail when any +ve number have even frequency then first one make it negative and second one positive(initial stage). So to solve it just make an extra check arr[index]>0 , only then make it negative(line no 13 in code ). so condition will be index0.
The following code works fine and passes all the test cases.
//Approach 2
//T.C -> O(N), S.C -> O(1)
//Uses concept of hashing
int missingNumber_two(int arr[], int n){
for(int i = 0; i < n; i++){
if(arr[i] 0){ //arr[index] > 0 to avoid considering the duplicate cases
arr[index] *= -1;
}
}
//In the below loop, if we find any element greater than 0, then that means (element's index + 1) element is not present in the array
for(int i = 0; i < n; i++){
if(arr[i] > 0){
return i + 1;
}
}
return n + 1; //If all the elements from 1 to n are present, then smallest element that is missing will be n + 1.
}
thanks
what about negative numbers ex :[-1,-2,-3,-4,-5] ?
☹️ 𝐩𝓻Ỗ𝓂Ø𝓈Ｍ
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playlist ke promotion se jada content pe dhiyan de dete jada like mil jaate or accha bhi lgta.....har video me bas playlist playlist
bakbass solution
Fails for [1,1], Dislike not work in leetcode
int firstMissingPositive(vector& nums) {
for(int i=0;i
check that value at that indes shloud ne positive then make it negative
AC SOLUTION
int n = nums.size();
// work on -ve and 0 element
for(int i = 0; i < n; i++) {
if(nums[i]
Wrong Solution
input
6
1 1 -2 4 5 2
Expected output
3
your Output
1