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You can use this logic to reduce the code. {
int count
for(int i = 0 ; i <= n; i++){
for(int j = 0; j < n; j++ ){ if(i < j){ if((a[i]+a[j])%k == 0){ count++; } } } }
}
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Divisible Sum Pairs
You are viewing a single comment's thread. Return to all comments →
You can use this logic to reduce the code. {
int count
for(int i = 0 ; i <= n; i++){
}