65 lines
1.5 KiB
C++
65 lines
1.5 KiB
C++
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/*
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@author: nandinisahu407
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special index-> if after deleting element from index i , sum of even index=sum of odd index
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approach->
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after deleting ,previous element at odd index will be now at even index and vice versa
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s_odd= odd[0 to i]+ even[i+1 to len]
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s_even=even[0 to i]+odd[i+1 to len]
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*/
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#include<iostream>
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using namespace std;
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int main(){
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int num;
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cout<<"enter length"<<endl;
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cin>>num;
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vector <int> arr (num);
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for(int i=0;i<num;i++){
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cin>>arr[i];
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}
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int count=0;
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int s_even,s_odd;
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for(int i=0;i<num;i++){ //checking whether special index or not
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s_even=0,s_odd=0;
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for(int j=0;j<i;j++){ // from index[0,i]
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if(j%2==0){
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s_even+=arr[j];
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}
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else{
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s_odd+=arr[j];
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}
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}
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for(int j=i+1;j<num;j++){ //from index[i+1,len]
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if(j%2==0){
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s_odd+=arr[j];
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}
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else{
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s_even+=arr[j];
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}
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}
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if(s_even==s_odd){ //checking whether sum of even index=sum of odd index
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cout<<"\n special index found at: "<<i;
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count++;
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}
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else{
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continue;
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}
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}
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cout<<"\n total special index: "<<count; //displaying
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return 0;
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}
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/* ----INPUT-----
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enter length 6
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4 3 2 7 6 -2
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----- OUTPUT------
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special index found at: 0
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special index found at: 2
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total special index:2
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*/
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//time complexity:o(n^2)
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