Build the prize algorithm from smaller procedures

Now connect the pieces. First compute the third distinct mark level in every subject once. Then call a Boolean procedure on each student: does at least one subject reach its cutoff? Only eligible students enter the total-score ranking.

Procedure SubjectTopper(X, Maths3, Physics3, Chemistry3)
    return (X.Mathematics >= Maths3 or
            X.Physics >= Physics3 or
            X.Chemistry >= Chemistry3)
End Procedure

Maths3 = TopThreeMarks(Scores, "Mathematics")
Physics3 = TopThreeMarks(Scores, "Physics")
Chemistry3 = TopThreeMarks(Scores, "Chemistry")
for each row X in Scores {
    if (SubjectTopper(X, Maths3, Physics3, Chemistry3)) {
        Insert X.Total and X.CardNo together into the best 3 slots
    }
}

A Boolean return is already a condition #

SubjectTopper returns True or False. if (SubjectTopper(...)) has the same effect as comparing that result with True. OR means at least one qualifying subject, not all subjects. A mark exactly equal to the cutoff qualifies because the comparison is >=.

Scores and IDs move together #

if (X.Total > FirstTotal) {
    ThirdTotal = SecondTotal; ThirdID = SecondID
    SecondTotal = FirstTotal; SecondID = FirstID
    FirstTotal = X.Total; FirstID = X.CardNo
}

Use analogous paired moves for insertion into second or third place. An ID of −1 can mark an empty prize slot because all real CardNo values are nonnegative. You could keep IDs alone and look up marks later, but keeping both avoids repeated lookups.

For these cards, the final paired slots are (281,11), (276,23), (261,18): Rahul, Deepika and Abirami. Our lab processes the same supplied records and shows eligibility separately from the running leaders. Subject cutoffs are saved before the ranking scan rather than recomputed inside every visit.

All three paired update cases #

FirstTotal = -1; SecondTotal = -1; ThirdTotal = -1
FirstID = -1; SecondID = -1; ThirdID = -1
for each row X in Scores {
    if (SubjectTopper(X, Maths3, Physics3, Chemistry3)) {
        v = X.Total
        if (v > FirstTotal) {
            ThirdTotal = SecondTotal; ThirdID = SecondID
            SecondTotal = FirstTotal; SecondID = FirstID
            FirstTotal = v; FirstID = X.CardNo
        } else if (FirstTotal > v and v > SecondTotal) {
            ThirdTotal = SecondTotal; ThirdID = SecondID
            SecondTotal = v; SecondID = X.CardNo
        } else if (SecondTotal > v and v > ThirdTotal) {
            ThirdTotal = v; ThirdID = X.CardNo
        } else {
            Record any equality with a retained score for tie review
        }
    }
}
Check the selected IDs, representation rule and any boundary ties
before declaring the final prize winners.

This expands the source’s strict-comparison update template. Equal totals do not create another distinct total level in that template. A general three-student award algorithm must retain tied candidates or apply its stated tie policy; do not assume these strict comparisons alone resolve ties. Our supplied final winning totals are distinct.