Coding-Decoding is one of those topics that looks deceptively simple until the exam throws a composite pattern at you and you spend 90 seconds staring at it. At its core, every coding question does exactly one thing: it transforms a word or number using a consistent rule, then asks you to apply that same rule to a new input.
Think of it like a vending machine. You put in "TIGER", a specific mechanism runs, and out comes "UJHDS". Your job is not to reverse-engineer the entire machine — it is to identify what the machine does in the first few seconds, then crank the handle yourself.
The "mechanism" almost always belongs to one of three families:
Letter-based codes — letters are shifted forward or backward in the alphabet by a fixed number (+1, +2, -3, etc.), or specific letters are substituted for other specific letters.
Position-based codes — each letter is replaced by its numerical position in the alphabet (A=1, B=2, ... Z=26). Sometimes the positions are reversed (A=26, B=25, ... Z=1).
Mixed or conditional codes — the shift alternates by position (odd letters shift by one rule, even letters by another), or the word is reversed before coding, or a combination of the above.
Here is the core analogy that makes everything click: treat the English alphabet as a number line from 1 to 26. Every coding rule is just an arithmetic operation on that number line — addition, subtraction, reflection, or a combination. Once you see it as arithmetic, you stop guessing and start calculating.
One critical mindset shift for IBPS PO: you will almost never need more than two example pairs to identify the rule. If you have cracked it in two pairs and it holds, trust it and move. Do not spend time triple-verifying — that is the clock-killer.
Every letter in the input word moves the same number of steps along the alphabet.
Rule: coded_letter = input_letter + n (where n is the shift, positive = forward, negative = backward)
Example: TIGER → UJHDS means T→U, I→J, G→H, E→F, R→S. Each letter shifts by +1.
How to detect it fast: check the first letter. If T→U, test the shift value (+1). Apply to the second letter. I+1=J? Yes. You are done diagnosing. Apply to the rest.
Watch the wrap-around: Z+1 = A, not a 27th letter. If W+3 = Z+1 = A, so W→Z, X→A, Y→B, Z→C.
Letters are replaced by their ordinal position number. This is extremely common in IBPS PO.
The table you must internalize:
A=1 B=2 C=3 D=4 E=5 F=6 G=7 H=8 I=9 J=10
K=11 L=12 M=13 N=14 O=15 P=16 Q=17 R=18 S=19 T=20
U=21 V=22 W=23 X=24 Y=25 Z=26
Detection: if the code is a sequence of numbers separated by hyphens or spaces, check the first letter-number pair. O=15 in ORANGE → 15. Confirmed. Apply to the rest.
Reverse position (A=26 ... Z=1) is less common but exists. If your first check gives a wrong answer, try 27 - standard_position.
The shift value depends on the position of the letter in the word.
Common variants:
Detection approach: after checking the first letter and finding the shift, check the second letter. If the second shift is different, you have a position-dependent pattern. Identify the shift for positions 1, 2, 3 and look for the alternating or sequential structure.
Example from the PYQs: BIG → CFJ, CAR → DDS, EAR → FDS. Check first letters: B→C (+1), C→D (+1), E→F (+1). Check second letters: I→J would be +1 but the code shows F (not J). Actually A→D is +3. Third letters: G→H (+1? No, the code shows J... wait). Re-read carefully — CFJ means C,F,J. B→C (+1), I→F... that is -3. Hmm. But CAR→DDS: A→D (+3). So the middle letter rule is +3, not -3. The issue is that "CFJ" has J for G, which is +3. So: position 1 shifts +1, position 2 shifts +3, position 3 shifts +1. DOG: D+1=E, O+3=R, G+1=H → ERH. But the answer is EPH. Re-examine: D+1=E (correct), O+1=P (shift +1 for even), G+1=H. So actually even positions shift +1 and middle position shifts +1 also... The point is: when in doubt, map each position individually across all given examples. Do not assume after one pair — confirm with a second pair.
The word is first reversed, then a shift is applied (or vice versa). Less frequent but appears occasionally.
Detection: decode the first example letter by letter. If the pattern seems inconsistent forward, try reading the coded word backward.
When you see a new coding question:
This entire diagnostic should take under 20 seconds on a well-internalized topic.
Memorize the five "anchor" letters E=5, J=10, O=15, T=20, Y=25. These divide the alphabet into clean segments. To find any letter's position: find the nearest anchor and count the offset. Example: R is 3 after O(15), so R=18. S is 4 after O, so S=19. This gets you any position in under 3 seconds versus writing out the full alphabet (15-20 seconds standard).
When a question uses reverse position coding (A=26, B=25...), use the formula: reverse_position = 27 - standard_position. A: 27-1=26. Z: 27-26=1. M: 27-13=14. You no longer need to re-count from Z downward. Standard method: count backward from Z (10+ seconds for letters in the middle). Shortcut: one subtraction (2 seconds).
In any letter-shift problem, check exactly two letter pairs from the first example. If both confirm the same shift, stop diagnosing and start applying. Most aspirants check all 5-6 letters of the example word — that is 3× the work needed. Two confirmed letters is sufficient. Time saved per question: approximately 15-20 seconds, which across 3-4 coding questions in a set = nearly a full minute recovered.
For position-based number codes, do not decode every letter of the answer options. Instead, decode only the first and last number of each option. Most IBPS PO distractors change exactly one digit — the first or last. Spotting which option has the correct first and last number eliminates 2-3 wrong options instantly. Full decode of the answer: 25-30 seconds. First+last spot-check: 8-10 seconds.
Before applying a forward shift to letters near the end of the alphabet (V, W, X, Y, Z), pre-flag them. V=22, so V+5=27=A (since 27-26=1, wraps to A). W+3=Z (no wrap, 25 is not past 26). The rule: if position + shift > 26, subtract 26 from the result to get the wrapped letter. Pre-flagging takes 1 second of awareness. Missing the wrap and choosing a wrong option costs the negative mark — the single most common mechanical error in this topic.
Here is the exact decision tree for the exam hall. Burn this in.
Step 1 — Identify code type (5 seconds) Are the codes letters or numbers? Numbers → almost certainly position code (A=1 to Z=26). Letters → shift or substitution.
Step 2 — Check first letter of Example 1 (5 seconds) Compute the relationship (shift value, position value). Note it.
Step 3 — Verify on second letter of Example 1 (5 seconds) Same relationship? Yes → fixed rule confirmed. No → note the second shift. Check for alternation pattern using positions 3 and 4.
Step 4 — Cross-verify on Example 2 if still unsure (10 seconds) Apply your hypothesized rule to Example 2's input. Does it produce Example 2's code? If yes → locked. If no → revise.
Step 5 — Apply to target word (10-15 seconds) Go letter by letter. Watch for wrap-around (Z boundary). Use EJOTY anchors for position lookups.
Step 6 — Spot-check answer options (5 seconds) Verify first and last character of your result against the options. Do not re-derive — match and move.
Total target time per question: 40-45 seconds. If you are beyond 60 seconds, mark your best guess and move — the question is costing more than it is worth.
Why this question: Tests the most fundamental IBPS PO coding type — straight alphabetical position. If you panic on number codes, this shows you there is nothing to panic about.
Solving path: Recognize the hyphenated number format immediately as position code. Verify: F=6 (6th letter), L=12 (12th letter). Confirmed. For PLANTS: P(16)-L(12)-A(1)-N(14)-T(20)-S(19). Use EJOTY: P is 1 after O(15)=16. N is 1 before O(15)=14. T=20 (anchor). S is 1 before T(20)=19. No arithmetic — pure anchor-counting. Time: under 25 seconds.
Why this question: The classic +1 letter shift. This is the single most common pattern on IBPS PO. You must crack it in under 10 seconds of diagnosis.
Solving path: T→U (+1), I→J (+1). Two checks — rule confirmed. ELEPHANT: E→F, L→M, E→F, P→Q, H→I, A→B, N→O, T→U = FMFQIBOU. Match option A. Total time: 30 seconds including option check.
Why this question: Same position-code logic as Q1 but with spaces instead of hyphens. Tests whether you recognize the pattern regardless of delimiter format.
Solving path: O=15, R=18. Confirmed position code. YELLOW: Y(25)-E(5)-L(12)-L(12)-O(15)-W(23). Y=25 (anchor: Y is itself an EJOTY letter). W: T(20)+3=23. Answer: 25 5 12 12 15 23. Option A.
Why this question: The +2 shift with a distracting first example (MOTHER→PQVJGT) that seems to give a +3 shift. This is a classic deliberate misdirection — the second example FATHER→HCVJGT resolves it cleanly.
Solving path: Check MOTHER: M→P is +3, but O→Q is +2 — inconsistency. Do not spiral. Jump to FATHER: F→H (+2), A→C (+2), T→V (+2), H→J (+2), E→G (+2), R→T (+2). Clean +2 throughout. MOTHER example has an error in the question — ignore it, trust FATHER. SISTER: S→U, I→K, S→U, T→V, E→G, R→T = UKUVGT. Option A.
Why this question: Tests position-dependent (mixed) shift where different letter positions in the word obey different rules. The most cognitively demanding type.
Solving path: BIG→CFJ: B→C (+1), I→F is wrong for +1. Try: I is position 2. CAR→DDS: A→D (+3) for position 2. EAR→FDS: A→D (+3) again for position 2. So position 2 shift = +3. Now position 3: G→H (+1), R→S (+1), R→S (+1). Position 3 shift = +1. Rule: position 1 = +1, position 2 = +3, position 3 = +1. DOG: D+1=E, O+3=R... but answer is EPH, meaning O+1=P. Re-examine: I→J for BIG is +1 (J is the 10th letter, I is the 9th). CFJ: C=3, F=6, J=10. B=2+1=3=C. I=9+1=10=J (J, not F). G=7+1=8=H. But the code says CFJ not CHJ. F=6, not J=10. There is an inconsistency in the given example. Trust the two clean examples (CAR and EAR) which both give position-2 shift = +3. But DOG's answer EPH means O→P (+1). The consistent pattern across all three examples for positions 1 and 3 is +1. For position 2: the two clean examples give +3. EPH implies O+1=P for DOG, suggesting +1 everywhere. Given the answer confirmed as EPH, apply uniform +1: D→E, O→P, G→H = EPH.
Why this question: Tests the +1 shift on a longer word, confirming you can sustain pattern application across 6+ letters without error.
Solving path: PENCIL→QFODJM: P→Q (+1), E→F (+1). Confirmed. ERASER: E→F, R→S, A→B, S→T, E→F, R→S = FSBTFS. Check wrap-around: none needed here (no late-alphabet letters). Option A.
Checking only one letter to confirm the pattern. One confirmation is not enough when position-dependent shifts exist. Always verify a second letter before applying the rule across the target word.
Missing the wrap-around at Z. Y+2 = A, not some imaginary 27th or 28th letter. Pre-flag any input letter from V onward when the shift is 3 or more. This is the most common single-mark loss in this topic.
Trusting a broken first example over a clean second example. Setters sometimes include a typographical error in the first given pair. If your rule perfectly explains Example 2 but not Example 1, trust Example 2 and cross-check your reading of Example 1.
Confusing forward shift with position code. A+1 shift gives B (still a letter). A position code gives 1 (a number). The code format — letters vs. numbers — tells you which family you are in before you do any calculation. Read the code format first.
Spending time on option elimination when direct application is faster. For a 4-option question where three distractors differ by only one digit, calculate the correct code directly. Do not try to eliminate options by reasoning — you will make mental errors under pressure.
Not using the EJOTY anchor system. Writing out A=1, B=2, C=3... in the margin is a 20-second drain every time you need a position value. The five EJOTY anchors (E=5, J=10, O=15, T=20, Y=25) get you any letter's position in 2-3 seconds via offset counting. If you have not internalized these five numbers, do it before your next mock.